manual for advanced physical chemistry laboratory-i
TRANSCRIPT
i
Manual for Advanced Physical Chemistry Laboratory-I
(BS 7th Semester & MSc 3rd Semester)
CODE No: 2568 UNITS: 1–9
By: Dr. Nasima Arshad
Assistant Professor
ALLAMA IQBAL OPEN UNIVERSITY
ISLAMABAD
ii
INTRODUCTION
This Physical Chemistry practical manual is designed for BS/MSc students studying in 7th
Semester – 3rd
Semester under 3 credit course with Course Code 2568. It consists of various
physical chemistry practicals. These practicals are divided into 09 units. The whole practical
course will be conducted in a 20-days workshop in the department of Chemistry, Allama Iqbal
Open University, Islamabad. Participation in the workshop practical work is mandatory for
students doing specialization in Physical Chemistry with at least 70% attendance and minimum
50% marks are required to pass this course.
Feel free to give your feedback, which will help us for the further improvement of this manual.
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COURSE OUTLINES
Unit 1: Extraction of DNA and determination of its purity using UV spectroscopy.
Unit 2: Determination of pKa of an indicator using Spectrophotometry.
Unit 3: Determination of ferric ions-salicylic acid complex composition, equilibrium
constant (Kc) and free energy change (ΔG) absorptometrically, using job’s
method.
Unit 4: Determination of partial molar quantities.
Unit 5: Determination of redox behavior by cyclic voltammentry and determination of
electrochemical parameters.
Unit 6: Determination of the rate constant for the acid-catalyzed hydrolysis of methyl
acetate and free energy change (ΔG) using titrometric method.
Unit 7: Iodination of cyclohexene and determination of rate constant absorptometrically.
Unit 8: Determination of molecular weight of an organic sample using Rast’s micro-
method.
Unit 9: Study of saponification of ethyl acetate with sodium hydroxide at equal
concentrations of ester and alkali and determination of rate constant (K) using
integrated rate law.
iv
RECOMMENDED BOOKS
1. C.H. Hamann, A. Hamnett, W. Vielstich, “Electrochemistry”, 2nd
edition, Wiley- VCH, 2007.
2. K.L. Chugh, “Modern Practical Chemistry”, H.P.U. edition,
Kalyani Publishers, New Delhi, 2007.
3. V.K.I. Ahluwalia, S. Dhingra, A. Gulati, “Practical Chemistry”,
Universities Press, India, 2005.
4. C.W. Shoemaker, G.J.W. Nibler and G.D. Christian, “Analytical
Chemistry”, 6th
edition, 2004.
5. C.W. Shoemaker, D.P. Garland and G.J.W. Nibler, “Experiments in
Physical Chemistry”, McGraw Hills, New York, 1989.
6. J. Sambrook, E.F. Fritsch, T. Maniatis, “Molecular Cloning: A
Laboratory Manual”, Cold Spring Harbor, New York, 1989.
7. P. David, “Experiments in Physical Chemistry”, 5th
edition, 1989.
8. A.M. James and F.E. Prichard, “Practical Physical Chemistry”, 3rd
edition, 1974.
9. A. Hussain, “Practical Manual”, Department of Chemistry,
University of Bahrain, Bahrain, Google Search.
10. B. Viswanathan, P.S. Raghvan, “Practical Physical Chemistry”,
Viva Books Private Limited, India, 2005.
1
Unit–1
DNA EXTRACTION
AND PURITY PROTOCOLS
2
EXPERIMENT 1
EXTRACTION OF DNA FROM CHICKEN BLOOD
GENOMIC DNA EXTRACTION Extraction of DNA basically consists of four major steps.
Preparation of a cell extract
Purification of DNA from cell extract
Concentration of DNA samples
Measurement of purity and DNA concentration
REAGENTS Buffer A (Red blood cell lysis buffer) composition
0.32 M sucrose
10 mM Tris HCl
5 mM MgCl2
0.75% Triton-X-100
Adjust pH to 7.6
Buffer B (Proteinase K buffer) composition
20 mM Tris-HCl
4 mM Na2EDTA
100 mM NaCl
Adjust pH to 7.4
Note: All solutions should be sterile. Buffer A should be autoclaved prior to addition of Triton-X-
100. Sterile filtering of solutions instead of autoclaving is a better option.
PROCEDURE 1. Add 1 volume of buffer A to 1 volume of blood and 2 volumes of cold, sterile, distilled, deionised
water. Vortex gently or invert tube 6-8 times and leave to incubate on ice for 2-3 minutes.
2. Spin at 3500 rpm for 15 minutes at 4oC. Discard supernatant into 2.5% bleach solution and re-
suspend pellet (vortex for 30 seconds at medium speed) in 2 ml of buffer A and 6 ml of water. Spin
at 3500 rpm for 15 minutes at 4oC. The pellet should be white to cream in colour. If pellet is
significantly red, repeat washing step again.
3. Add 5 ml of Buffer B and 500 µl of 10% SDS to pellet. Re-suspend pellet by vortexing vigorously
for 30-60 seconds. Then add 50 µl of Proteinase K solution (20mg/ml). The Proteinase K solution
should be made fresh and refrigerated prior to use.
4. Leave to incubate for two hours at 55oC in a water bath. Remove samples and leave to cool to room
temperature (or leave for 2-3 minutes on ice). Add 4 ml of 5.3 M NaCl solution. Vortex gently for
15 seconds.
5. Spin at 4500 rpm for 15-20 minutes at 4oC. Pour off supernatant into a fresh tube. Take care not to
dislodge pellet. Add an equal volume of cold isopropanol (stored at -20oC). Invert 5-6 times gently
to precipitate DNA.
6. Remove DNA with a wide bore tip and transfer to a microfuge tube. Wash with 1 ml of 70%
ethanol. Leave DNA to dry for 15-20 minutes at 37oC. Re-suspend in 300-400 µl of Tris HCl, pH
8.5 (not TE!). Leave to re-dissolve overnight at room temperature. DNA can be safely refrigerated
for up to a year. Long-term storage may involve ethanol at -70oC.
3
EXPERIMENT 2
PROTOCOL FOR DNA EXTRACTION FROM CALF THYMUS GLAND
PURPOSE In this lab protocol, you are going to extract the DNA from thymus gland nuclei. The thymus is a gland
that is very large in immature mammals. It serves a function as a part of the immune system so there are
many white blood cells with many large nuclei. Many thousands of thymus cells will be used for the
extraction, so you will be combining the DNA from thousands of nuclei. In this way, you should be able
to see long, combined strands of DNA.
MATERIALS FOR PREPARATION
Test tubes, 16 Test tube racks, 16 Clinical centrifuge
Prep buffer NaCl Solution EDTA Solution
SDS Solution Ice cold 95% ethanol Small cups or beakers for solutions
Thymus Blender 15 ml cap-less centrifuge tubes, 8
Hand Strainer 250 ml beakers Graduated cylinder, for measuring
Glass stir rods Freezer Refrigerator
PROCEDURE
1. ISOLATE NUCLEI This step will be done earlier, or as a demonstration.
a. Puree 10 g thymus gland with ~10 ml prep buffer in blender
b. Mix pureed thymus with 200 ml prep buffer
c. Stir until well blended.
d. Strain to remove large chunks of tissue – save liquid & toss solids.
(Note: a – d can be done up to 24 hours ahead of time. Store in the refrigerator)
e. Centrifuge 4 ml of strained liquid for 5 minutes. This makes a pellet of the nuclei. (nuclei are the
heaviest part of the cell.)
f. Pour off supernatant (the liquid on top of the pellet).
g. Resuspend pellet in 2 ml (~44 drops) prep buffer. Mix well with plastic beral pipet. Make sure all of
pellet is dissolved.
h. Put ½ of solution into each of 2 test tubes, ~ 1 ml per tube. Resuspended pellet is shared by two lab
groups.
i. Prepare a slide with a drop of this solution; stain with methylene blue; observe with a microscope;
diagram and describe what you see in your notebooks. This may be unsuccessful if the thymus has
been frozen.
2. LYSE NUCLEI a. Add ½ ml (~11 drops) of EDTA solution (this binds Mg and Ca ions which are needed by enzymes
lurking in the cytoplasm from degrading the DNA as it is released from the nuclei). Mix gently.
b. Add 100 µl (~2 drops) of SDS solution (a biological detergent similar to shampoo which solubilizes
proteins and disorients fats in the cell membranes). Mix gently.
c. Add 250 µl (~5 drops) of NaCl solution, one drop at a time, mixing gently after each drop.
3. PRECIPITATE DNA
a. Gently add ~1 ml of ice cold 95% ethanol by pipetting slowly down the side of the test tube. The
4
alcohol will form an overlay. You should begin to see strings of DNA precipitating at the point of
the overlay and reaching up into the ethanol layer. (DNA precipitates because it is not soluble in
ethanol. All of the other components of the cell are soluble in the ethanol).
b. Gently spool DNA threads where the ethanol and DNA mixture meet. There should be gobs of
DNA. It is white, stringy and looks a bit like mucus.
PROTOCOL FOR SOLUTIONS AND REAGENTS
WORKING SOLUTIONS
Preparation Buffer Sodium Dodecyl Sulfate
(aka sodium laurel sulfate)
57.0 g Sucrose 25 g SDS
3.1 g MgCl2-6H2O distilled H2O to a final volume of 250 ml
0.6 g Tris.HCl Caution – this is a detergent, bubbles easily
400 ml distilled H2O Store at room temperature
Adjust to pH 7.5 w/ 0.1 N NaOH
Bring to a final volume of 500 ml Sodium Chloride
Store in refrigerator 29.2 g NaCl
Long term storage should be in freezer Distilled H2O to a final volume of 250 ml
Store at room temperature
Ethylene diamine tetra acetic acid Denatured 95% Ethanol 0.72 g EDTA Adjust to pH 7.5 w/ 0.1 N NaOH
200 ml distilled H2O Store in freezer (use ice cold)
Bring to a final volume of 250 ml Store at room temperature
HINTS 1. The proportion of thymus to prep buffer needs to be kept the same as in the original recipe. (10 g
thymus into 200 ml prep buffer).
2. The proportions of extract solution to SDS, NaCl. EDTA and ethanol need to be the same as in the
original recipe.
3. It is a good idea to add the prep buffer to the ground up thymus slowly. If the pink color begins to
disappear, do not add the full amount.
4. When too little DNA is available, you must observe quickly. You may be able to see white threads
floating up into the ethanol, but they disappear quickly, within 10 – 20 seconds.
5. The tip of the spooling rod should not go below the interface of the ethanol and the extract solution.
6. Make sure the thymus is being ground by the blender, sometimes the small chunks slide under the
blades, rather than through them. Your solution should be pinkish.
7. After sitting for a few minutes, a whitish solute should be seen dropping to the bottom of the
beaker. These are your nuclei.
8. If you do not have a centrifuge, you can let the solution sit for an hour or so (or overnight) and the
nuclei will drop to the bottom of the beaker. You then pour off the top part of the solution. Keep
refrigerated.
5
TROUBLESHOOTING
Problem Why Solution
1. Little to no DNA in tube
DNA is too dilute
Try spinning 8 ml of thymus in prep buffer instead of 4 ml. Use more thymus or fewer buffers. Stir the thymus solution before taking the 4 ml. (Because the nuclei drop, the top part of the solution becomes nuclei deficient).
2. Not spooling DNA not sticking to rod
Try a plastic pipette, coffee stirrer, or roughen up the sides of the glass rod. Swirl the rod around gently, and then run it up the side of the tube. Make sure that the tip of the spooling rod does not go into the aqueous phase.
3. Calf thymus solution is not pink
DNA is too dilute Grind up some more thymus and put into same buffer. Do not use any new buffer.
4. Do not need so much solution
Doing fewer experiments Use half as much thymus and half as much buffer. Do the rest of the experiment the same.
5. Ethanol does not float on top
Ethanol solution is too dilute
Use a higher percentage alcohol
6. Can’t spin solution
No centrifuge
Make the thymus solution a day early and let sit in the refrigerator overnight. Carefully remove beaker from refrigerator and pour off the top part. Do not pour out any of the solute.
7. Can’t weigh the thymus
No balance
10 g of thymus should be a little bigger than the size of your thumb. Add the prep buffer to the thymus solution slowly. If the pink starts to disappear, stop adding. Do not add more than the recipe says.
ANALYSIS 1. What is the purpose of centrifuging the thymus mixture after it has been pureed and strained?
2. Why do you need to bind Mg and Ca ions?
3. What common household chemicals contain SDS?
4. Why is a salt solution added?
5. What properties of DNA are responsible for its precipitation in alcohol?
6. If you looked at the spooled material under the microscope, would you be able to see the
nucleotides that are the building blocks of DNA?
7. Why might freezing the thymus alter the ability to see nuclear material in the cells (step 4)?
8. Compare the general structure of an animal cell with a plant cell and a bacterial cell.
9. a. What additional step would you have to take to extract the DNA of plant or bacterial cells?
b. What could you use to accomplish that process?
6
EXPERIMENT 3
DERERMINATION OF DNA CONCENTRATION
AND PURITY USING A SPECTROPHOTOMETER
Check purity and concentration of the DNA by measuring the absorbance at 260 and 280 nm.
DNA concentrations can be accurately measured by UV absorbance spectrometry. The amount of UV
radiation absorbed by a solution of DNA is directly proportional to the amount of DNA sample. Usually
absorbance is measured at 260 nm, at which wave length an absorbance of 1.0 corresponds to 50 µg of
double-stranded DNA per ml. UV absorbance can also be used to check the purity of a DNA preparation.
With a pure sample of DNA the ratio of the absorbancies at 260 nm and 280 nm (A260/A280) is 1.8. Ratios
of less than 1.8 indicate that the preparation is contaminated, either with protein or with phenol.
PROCEDURE 1. Turn on the spectrophotometer and the UV lamp. Set the wave length at 260 nm. Let warm up for at
least 5 min.
2. Prepare blank with 20 µl of EB Buffer mixed with 980 µl of sterile water.
3. Dilute your sample in the same ratio: 20 µl of EB Buffer mixed with 980 µl of sterile water.
4. Using UV-permeable cuvettes, zero the spectrophotometer with the blank solution. Measure the
absorbance of the diluted DNA preparation.
5. Change the wavelength to 280 nm.
6. Re-zero with the blank.
7. Take the absorbance of the diluted DNA preparation at 280 nm.
8. Calculate the purity and concentration of the DNA in your preparation.
HINT: 1 µg/ml = 0.001 µg/µl = 1 ng/µl.
Using units per µl will make it easier to calculate the amount of DNA to add in subsequent reactions.
If your DNA concentrations are too low, we will re-measure the concentrations using the Nanodrop
spectrophotometer. The Nano drop has a lower detection threshold and is able to obtain accurate readings
using only 1-2 µl of sample.
PRECUATIONS Properly label the tubes containing your genomic DNA and plasmid DNA. Store them in the fridge
at 4 °C.
7
Unit–2
STRENGTH OF AN ACID (PKa)
BY SPECTROSCOPIC METHOD
8
EXPERIMENT 1
SPECTROPHOTOMETRIC DETERMINATION
OF THE pKa OF AN ACID-BASE INDICATOR
BACKGROUND Colorful acid -base indicators are organic weak acids or bases that change color at different pH. In this
experiment, spectrophotometry is employed to measure the pKa of bromothymol blue, an acid-base
indicator. The indicator (HIn) is a monoprotic acid and we can represent its dissociation as follows:
HIn → H+ + In
- (1)
The equilibrium expression for such dissociation can be written as
pH = pKa + log ( [In-] / [HIn] ) (2)
This can be rearranged to a straight-line equation
log ( [In-] / [HIn] ) = pH – pKa (3)
Hence, if the log term is plotted vs. pH, the slope is 1, the intercept is -pKa and the line should cross the
pH axis at pH = pKa (Fig. 1 left).
The ratio of [In-] / [HIn] can be determined spectrophotometrically. First, a solution of bromothymol blue
is prepared in acidic solution where essentially the entire indicator remains in the HIn form. The
absorption spectrum is then determined. Second, a solution is prepared in the basic solution where the
indicator is in its basic form, In-. The absorption spectrum of In
- is then determined. From the two
absorption spectra the wavelengths of maximum absorbance of HIn and In- are selected for further
measurements.
Buffered solutions with pH values on either side of the pKa of the indicator are then prepared and the
absorbencies measured at selected wavelengths. The solutions contain the same total concentration of
indicator, [In-] + [HIn], but the ratios vary with pH. Fig.1 right shows a typical plot of absorbance vs. pH
at the wavelength of maximum absorbance for the In- species.
The terms used in this Fig. are as follows:
A = absorbance of mixture
Aa = absorbance of the weak acid, HIn
Ab = absorbance of the weak base, In
Fig.1 Left: Log plot of Eq. 3. Use the linear least square lines on your data to obtain the best pKa and its
error. Approximate pKa is shown on the graph. Right: Plot of absorbance vs. pH at the maximum
wavelength of absorbance for In- species.
9
From the graph it is evident that [In-] / [HIn] = (A - Aa) /(Ab - A). If the wavelength used is the one at
which HIn shows a maximum absorbance, the curve will be similar to that shown in Fig. 1 (Right), except
that it will start at high absorbance and curve down to a low absorbance value at high pH.
PROCEDURE
PART I PREPARATION OF SOLUTIONS
1. Weigh out precisely 0.02 g of bromothymol blue and dissolve in 200 mL of 20% ethanol.
2. Prepare 100 mL of 0.20 M NaH2PO4 • H2O (FW 137.99). Calculate actual molarity.
3. Prepare 100 mL of 0.20 M Na2HPO4 • 7 H2O (FW 268.07). Calculate actual molarity.
4. Prepare a small amount of 3 M NaOH. Three grams of NaOH dissolved in 25 mL of solution will
provide enough of this reagent.
5. Prepare a series of buffered solutions as follows: Secure nine clean 100 mL volumetric flasks (if it
is more convenient to use 50 mL or 25 mL volumetric flasks, the amounts of reagents should be
scaled down proportionately). Pipette 2.00 mL of the bromothymol blue solution into each
volumetric flask. Then add the following volumes of phosphate solutions to these volumetric flasks.
(These volumes can be measured with a 10 mL graduated cylinder).
Flask NaH2PO4 Na2HPO4
Number Volume (mL) Volume (mL)
1 0.0 0.0
2 5.0 0.0
3 5.0 1.0
4 10.0 5.0
5 5.0 10.0
6 1.0 5.0
7 1.0 10.0
8 0.0 5.0
9 0.0 0.0
The above nine solutions must be made in the proportions listed. If 50 mL or 25 mL volumetric
flasks are used, adjust the volumes proportionally.
To Flask 1 add 5 mL of DI water and 8 drops of 6 M HCl. To Flask 9 add 5 mL of DI water and 16
drops of 3 M NaOH solution. Now dilute each solution to the mark and mix thoroughly. Measure
and record the pH value of each solution using a pH meter.
6. Measure the absorbance of Flask 1 from 400 to 700 nm using water as a blank, and determine the
absorption spectrum of bromothymol blue at low pH (the spectrum of HIn). Read the absorbance at
20 nm intervals except in the vicinity of the maximum, there readings should be taken every 5 nm.
Plot the absorbance vs. wavelength.
7. To determine the absorption spectrum at high pH (the spectrum of In-), measure the absorbance of
the Flask 9 in the same manner as directed above. Plot the results on the same graph. Your plot
should look similar to the following.
10
8. Using YOUR spectra, select two wavelengths at which further absorbance measurements will be
made. Choose wavelengths at which HIn and In- exhibit maximal differences in absorbance.
9. Now measure the absorbencies of each of the nine solutions at the two wavelengths selected. To
organize your data in your notebook you should make the following table:
Flask Number Absorbance at λ1 Absorbance at λ2 pH Solution color
λ1 = nm λ2 = nm
1
2
3
4
5
6
7
8
9
Prepare a graph of absorbance vs. pH for each of the two wavelengths (like Fig.1 Right). Determine the
[In-]/[HIn] in solutions 2 to 8 as explained above and shown in Fig. 1 Right. Plot log([In
-]/[HIn]) vs. pH
(shown by Fig. 1 Left) and obtain a value of pKa from the best fit line. Also, calculate the measurement
error. Compare your result with the literature value and discuss the effect of ionic strength on the pKa
value. Discuss all other sources of errors.
QUESTIONS 1. Explain why it is not necessary to know the exact concentration of bromothymol blue in the
solution.
2. The pKa value can also be determined from Fig. 1Right - explain.
3. The acid dissociation constant for bromothymol blue is dependent on ionic strength of solution.
Determine the ionic strength of bromothymol blue by interpolating the values given below.
Ionic strength and associated pKa of bromothymol
μ pKa
0.01 7.19
0.05 7.13
0.1 7.1
0.5 6.9
11
EXPERIMENT 2
SPECTROSCOPIC DETERMINATION OF INDICATOR PKa
BACKGROUND pH indicators may be defined as highly colored Bronsted-Lowry acid-base conjugate pairs. Used in low
concentrations, these compounds signal pH changes within a specific range determined by the particular
indicator in use. This color change range depends upon the relative acid strength (or pKa ) of the
conjugate acid form of the indicator. In many indicator systems, both conjugate species are colored, and,
within the pH transition range, the observed color is really a mixture of the colors of the two forms.
The ratio of concentrations of the conjugate acid and base forms is controlled by the pH of the solution, as
indicated by the same equation as that used for determining buffer pH’s, namely, the Henderson-
Hasselbalch equation:
log [conjugate base] / [conjugate acid] = log [Ind-]/[HInd] = pH - pKa
Where Hind and Ind- represent the acidic and anionic forms of the indicator, respectively.
The indicator concentration ratio is controlled by the pH of the solution, whereas the buffer concentration
ratio controls the solution pH. At first glance, this statement seems contradictory. In reality, however, it is
simply a matter of relative concentrations. The buffer components are present in high concentrations, so
they control the pH of the buffer solution via the conventional acid-base reactions. On the other hand, the
indicator species are present in low, even negligible, amounts relative to the other acid-base systems in
the solution. In terms of visible absorption, however, the indicator species predominate over the buffer
components (most of which are colorless).
Procedures for determining solution pH using a short range of visual comparisons of indicator colors can
usually distinguish 0.2 pH unit differences over a pH range of 2.0 units. This analytical approach is improved
if a spectroscopic instrument is used to accurately measure absorbance of one or both of the indicator forms.
To adequately interpret such data, knowledge of the value of the indicator pKa is essential.
In the present experimental procedure, previously obtained knowledge of the acetate buffer system and
the Henderson-Hasselbalch equation will be used to determine pKa values of various indicators. As in the
earlier buffer experiment, standard solutions of 1.0 M acetic acid and 1.0 M sodium acetate will be mixed
to produce buffer solutions of various pH’s. This time, however, the buffer pH will be calculated from the
Henderson-Hasselbalch equation, using the previously determined pKa value for the acetic acid-acetate
ion system.
An additional modification of the procedure is that a constant total concentration of indicator will be
added to each buffer mixture. This operation involves the most critical measurement of the experiment;
unless equal total amounts of indicator are present in all buffers, the spectroscopic measurements will be
meaningless. A number of indicators are provided for analysis, as listed in Table 1 (along with pertinent
spectroscopic information).
Table 1
Indicator (concentration) acid form (λmax) base form (λmax)
Bromocresol green (67 mg/l) yellow (453 nm) blue (610 nm)
Chlorophenol red (33 mg/l) yellow (433 nm) magenta (575 nm)
Bromocresol purple (40 mg/l) yellow (432 nm) purple (590 nm)
12
To simplify calculations, all spectroscopic measurements will be made at the wavelength of maximum
absorbance, λ max, of the base form of the indicator. Each student (or group of students) will make three
determinations of the pKa of one of the indicators, each determination to be made with a different buffer
mixture (and buffer pH value). The absorbance of the indicator in unmixed (or pure 1.0 M) acid solution
will be measured to determine the minimum absorbance level of the base form, designated “Aa “, and the
absorbance of indicator in unmixed (or pure 1.0 M) salt solution is measured as the maximum absorbance
of the base form, designated “Ab”.
The spectroscopic determination of indicator pKa , involves calculations based on the following argument.
The total indicator concentration is the same for all buffer mixtures and is proportional to the value (Ab -
Aa) if all measurements are made at the λ max of the base form. In each buffer, the indicator is distributed
between two forms, acid and base, the relative amounts of each determined by the buffer solution pH. The
concentration of the base form, [Ind ─], is proportional to (Ai - Aa) where “Ai” is the absorbance of the
particular buffer sample under study.
The indicator acid form concentration, [HIind], is then proportionate to that part of the total amount not in
the base form. Thus, the [HInd] is given by the relationship:
(Ab - Aa) - (Ai - Aa) = (Ab - Ai). Since the proportionality constant for these concentration relationships is
the same, it cancels out when a ratio is made of the concentration terms. Thus, neither total indicator
concentration nor any “ab” term from absorbance values appears in the calculations. The ratio [Ind-
]/(HInd] equals the ratio of absorbance terms or
(Ai - Aa)/(Ab - Ai).
The indicator pKa is calculated by substituting the absorbance ratio term (for the ratio of salt/acid) and a
theoretical buffer pH value into a modification of the Henderson - Hasselbalch equation, as:
[ ]log
[ ]
i aa
b i
A ApK pH
A A
PREPARATION OF INDICATOR / BUFFER MIXTURES Working in groups of not more than three students, prepare the indicator/buffer mixtures in the familiar
manner. Three burets will be involved in the operation: the first contains 1.0 M acetic acid, the second 1.0
M sodium acetate, and the third the indicator of choice.
Mix the reagents as specified in Table 2. Note that 2.00 mL of indicator is added to each solution (buffer
mixtures, pure acid, and pure salt alike).
13
Table 2
Sample # mL Hind mL salt mL acid Salt/acid ratio Long ratio Buffer pH
1 2.00 0.00 8.00 -------- -------- --------
2 2.00 2.00 6.00
3 2.00 4.00 4.00
4 2.00 6.00 2.00
5 2.00 8.00 0.00 -------- --------
Calculate the pH of each buffer mixture using the original Henderson-Hasselbalch equation:
pH = pKa+ log ([salt] / [acid]). The pKa in this equation is for the dissociation of acetic acid as
determined in a previous experiment (it should have a value of about 4.62). Notice that the dilution effect
from adding the indicator cancels in the buffer salt/acid ratio.
MEASUREMENT OF INDICATOR / BUFFER ABSORBANCES
Turn on a Spectronic 20, and allow the instrument to warm-up. Set the wavelength dial to λ max for the base
form of the indicator chosen for study. Place each of the indicator/buffer solutions in a clean, dry, or properly
rinsed cuvette. Use deionized water in a sixth cuvette to calibrate the spectrometer to read 100% T.
Record the observed %T (to ±0.1 %) and A for each of the samples in data Table 3, and calculate Acalc
using the equation: A = 2 - log %T
Table 3
Sample # %T. Ameas Acalc
1 = Aa
2
3
4
5 = Ab
Note the designations given to the absorbance values for Samples #1 and #5. The value for Sample #1 is
Aa, the minimum absorbance measured at λ max for the base form of the indicator (since the indicator is
in a solution of pure 1.0 M acetic acid). Correspondingly, the absorbance of sample #5 is Ab, the
maximum absorbance of the base form (since the indicator is in 1.0 M salt solution).
DETERMINATION OF INDICATOR pKa
Use the absorbance values recorded in Table 3 to calculate the indicator pKa’s for the buffer mixtures
(Samples #2, #3, and #4). Complete the entry columns in Table 4, and calculate the indicator pKa for the
modified Henderson-Hasselbalch equation:
[ ]log
[ ]
i aa
b i
A ApK pH
A A
Table 4
Sample # (Ai-Aa) (Ab-Ai) [Ai - Aa]/ [Ai- Aa] Log ratio Hind pKa
2
3
4
14
DATA TREATMENT AND REPORT
A two-page report is required for this experiment. On the first page, under appropriate headings, make
complete copies of Tables 2, 3, and 4. List the name (and pertinent spectroscopic data) of the indicator
used in the experiment, and then give the calculated pKa for the indicator system.
On the second page of the reports answer the following questions, giving a clearly thought-out
explanation of each answer.
1. What single error would have the greatest effect on the accuracy of the experimental results?
2. All indicator pKa values in this experiment are within 2 units of the pKa of acetic acid. Is this
necessary to the method, or can any indicator pKa be determined in acetate buffer solutions?
15
Unit–3
COMPLEX COMPOSITION
BY JOB’S METHOD
16
EXPERIMENT 1
DETERMINATION OF FERRIC IONS-SALICYLIC ACID COMPLEX
COMPOSITION, EQUILIBRIUM CONSTANT (KC) AND FREE ENERGY CHANGE
(ΔG) ABSORPTOMETRICALLY, USING JOBS METHOD.
BACKGROUND The weak acid 2-hydroxybenzoic acid, traditionally known as salicylic acid makes coordination complex
with iron in solution. The concentration of the iron complex may be measured spectrophotometrically. In
acid, a violet complex is formed. At neutral pH, a different dark-red complex forms, and in basic solution
the complex that is formed is orange. This experiment will be conducted at acidic pH. This complex is
stable in acidic olution in pH range 2.6 to 2.8. This range can be obtained by using solutions in 0.002M
HCl. At this pH value the phenolic –OH and –COOH group of the acid remin unionized. Since the copex
is colored, its concentration can be measured on an ordinary spectrophotometer in visible region. In any
complex formation the bond units combine in integral numbers.
According to Job’s method, when equimolar solutions of two reactants are mixed in varying proportions,
at equilibrium the maximum amount of the complex is formed when the two reactants are present in
samemole ration as is required for the complex formation. When the complex is colored (in the present
case the complex formed is deep violet in color)and the reactants are colorless, the equilibrium
concentration of the complex can be followed by measuring absorbance of the solution.
OBJECTIVE The main objective of this experiment is to determine ferric ions-salicylic acid complex
APPARATUS Spectrophotometer, test tubes, beakers, measuring flasks etc.
CHEMICALS/SOLVENTS Salicylic acid, ferric alum. HCl
PROCEDURE
1. Prepare 0.002M HCl solution as stock and then prepare 0.001M salicylic acid (mol wt.=138g) and
0.001M Fe+3
ions as ferric alum each in 0.002M solution.
2. Mix both solutions in following compositions
Ferric ion soln (ml) 9 8 7 6 5 4 3 2 1
Salicylic acid soln (ml) 1 2 3 4 5 6 7 8 9
3. Take any of these solutions and measure the maximum absorption and also wavelength of light
transmitted (λmax). Measure the absorption for each of the above solutions at this wavelength.
4. Plot absorption vs. volume of Fe3+
ion solution or mole fraction. Maximum point on the curve
corresponds th the composition of the complex (job’s plot).
5. Now determine the molar absorptivity coefficient (ε) of the complex. For that pipette ou 1,2,3 and
4mL of 0.002M Fe3+
ion solution into four stoppered test tubes. To each test tube add 15 mL of
saturated salicylic acid solution and enough 0.002M HCl solution to bring the total volume to
10mL. Measure the absorbance of each solution. Plot absorbance vs. [Fe3+
]. (Because the salicylic
acid is in excess, it is assumed that the concentration of iron equals the concentration of the
17
complex. Calculate ε by linear regression (Beer-Lambert Law (A= εcl, where 1=1cm {path length
of cell}).
CALCULATIONS
1. Tabulate the absorbance data
2. Draw Job’s graph by plotting absorption vs. volume of Fe3+
ion solution or mole fraction and find
the maximum point on the curve corresponding to the equilibrium composition of the complex
(job’s plot).
3. Plot absorbance vs. [Fe3+
] as mentioned in the procedure and find molar absorptivity coefficient
(Lmol-1
cm-1
).
4. Using the value of molar absorptiviy coefficient (ε) and absorbance (at maximu point on Jab’s
curve), find the concentration of complex (equilibrium mixture) as A= εcl.
5. Euilibrium constant of the complex is less than calculated by using following equation;
Kc = [FeC6H4(OH)COOH]3+
/ [Fe3+
] [C6H4(OH)COOH]
Where,
[FeC6H4(OH)COOH]3+
= Equilibrium concentration of the complex
[Fe3+
] = Concentration of iron used at equilibrium
C6H4(OH)COOH = Concentration of salicylic acid used at equilibrium
6. From equilibrium constant data, calculate Gibbs free energy change as ΔG = -RT ln Kc KJ mol-1
18
EXPERIMENT 2
METHODS FOR DETERMINING THE
COMPOSITION OF COMPLEXES IN SOLUTION
BACKGROUND Complexation reactions of the form
xM + yL ↔
_ MxLy (1)
are based on the reaction of a metal cation (M) and a ligand (L). These reactions are widely used in
analytical chemistry. Absorption spectroscopy is a powerful tool for exploring these complexation
reactions. In this experiment, two general approaches to studying the composition of complexes are used
to demonstrate the necessity of carefully evaluating the properties of a particular chemical system in order
to select the best method for determining the composition (metal to ligand ratio) of a complex by
absorption measurements.
Fig. 1. Continuous-variation plot for a 1:2 complex, ML2.
METHOD OF CONTINUOUS VARIATIONS (JOB'S METHOD) In this method, metal cation and ligand solutions with identical concentrations are mixed in different
amounts such that the total volume of the mixture solutions and the total moles of reactants in each
mixture is constant. This procedure causes the mole ratio of reactants to be varied across the set of
mixture solutions. The absorbance of each solution is then measured and plotted vs. the volume fraction
of one of the reactants (M or L). For example, the volume fraction of the metal is
VM/(VM + VL) (2)
where VM is the volume of the metal cation solution and VL is the volume of the ligand solution. A plot of
this type is reproduced above in Fig. 1.
19
Assuming the complex absorbs more than the reactants, a maximum occurs at a volume ratio VM/VL
corresponding to the combining ratio of cation and ligand in the complex. At other volume ratios, one of
the reactants is a limiting reagent. In the plot in Figure 1, this maximum occurs at VM/VL = 0.33/0.66,
suggesting a complex of the formula ML2.
Fig. 2. Mole-ratio plots for 1:1 and 1:2 complexes.
20
MOLE-RATIO METHOD (YOE-JONES METHOD) In this method, a series of solutions is prepared in which the concentration of one reactant is held constant
while that of the other is varied. The absorbance of each solution is measured and plotted vs. the mole
ratio of the reactants. Assuming the complex absorbs more than the reactants, this plot will produce an
increasing absorbance up to the combining ratio. At this point, further addition of reactant will produce
less increase in absorbance. Thus a break in the slope of the curve occurs at the mole ratio corresponding
to the combining ratio of the complex. A sample mole-ratio plot is included in Fig. 2 above.
REAGENTS
Ferrous ammonium sulfate ((NH4)2Fe(SO4)2⋅6H2O, FW = 392.13), 7.0 x 10-4
M
(250 mL) 1,10-phenanthroline (FW = 198.23), 7.0 x 10-4
(250 mL) M.
Acetic acid (17.45 M) / sodium acetate (FW = 136.08) buffer, pH 4.0, total acetate = 0.01 M
Hydroxylamine hydrochloride (FW = 69.49), 5 % (w/v = g/mL) (50 mL).
PROCEDURE
SOLUTION PREPARATION Prepare the solutions listed above in the specified amounts. Make the phenanthroline solutions first. To
prepare the buffer, make 250 mL of 0.01 M sodium acetate and titrate to pH 4.0 by addition of acetic acid
(monitor addition with pH meter). Prepare a spectroscopic blank by adding 5 mL of the acetate buffer and
1 mL of the hydroxylamine hydrochloride solution to a 25 mL volumetric flask and diluting to the mark
with distilled water.
ABSORBANCE SPECTRA OF STOCK SOLUTIONS Record the absorbance spectrum of each of your stock solutions vs. a distilled water reference.
METHOD OF CONTINUOUS VARIATIONS Transfer 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 mL of the standard Fe
2+ solution to separate 25 mL volumetric
flasks. Add 5 mL of the acetate buffer solution followed by 1 mL of the hydroxylamine hydrochloride
solution. To each flask, respectively, add 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and 0 mL of the 7.0 x 10-4
M 1,10-
phenanthroline solution. Dilute to the mark with distilled water and mix. After ten minutes, record an
absorbance spectrum of the 5mL Fe2+/
5ml 1,10-phenanthroline solution (vs. your spectroscopic blank)
and determine the wavelength of maximum absorbance of the complex. Evaluate the spectra of the stock
solutions to find a wavelength for the complex that has a maximum absorbance while possessing
minimum overlap with any absorption bands due to the chemical species present in the stock solutions.
Use this wavelength for all further absorbance measurements. Measure the absorbance of each solution in
triplicate.
Mole-Ratio Method. Transfer exactly 2 mL of the standard Fe2+
solution to eight separate 25 mL
volumetric flasks. Add 5 mL of the acetate buffer solution followed by 1 mL of the hydroxylamine
hydrochloride solution. Add 2, 3, 4, 5, 6, 8, 10, and 12 mL of the 7.0 x 10-4
M 1,10-phenanthroline
solution to the various flasks. Dilute to the mark with distilled water, and mix. After ten minutes measure
the absorbance of each solution vs. the spectroscopic blank. Make three replicate absorbance
measurements.
CALCULATIONS Sections 1 and 2 below produce plots with linear portions that intersect. For each plot, associate each
point with either the left or right linear portions and perform two least-squares calculations to estimate the
linear equations (y = mx + b) that represent the left and right linear portions of the plot. Slopes, intercepts,
and values of r2 should be tabulated for each of these least-squares.calculations.
21
The intersections of the lines can be found mathematically by finding the point on the x axis where the
lines have the same y -value (i.e., setting the equations equal to each other and solving for x). If replicate
absorbance measurements have been made, all plots should have error bars based on the 95% confidence
intervals of the absorbance measurements.
For the method of continuous variations, plot absorbance vs. mole fraction of Fe2+
. Extrapolate the linear
portions of the curve and locate the intersection by use of the mathematical procedure described above.
For the mole-ratio method, plot absorbance vs. moles of reagent/mole of iron. Extrapolate the linear
portions of the graph and locate the intersection by use of the mathematical procedure described above.
For the mole-ratio method, plot absorbance vs. moles of reagent/mole of iron. Extrapolate the linear
portions of the graph and locate the intersection by use of the mathematical procedure described above.
Postulate the formula for the complex and draw its structure. In the discussion section of your report,
comment on any discrepancies between the combining ratios suggested by the two measurement
approaches.
22
23
Unit–4
PARTIAL MOLAR QUANTITIES
24
EXPERIMENT
DETERMINATION OF PARTIAL MOLAR PROPERTIES OF SOLUTIONS
A solution is a homogeneous mixture forming a one phase system with more than one component.
Just as the behavior of gases is discussed in terms of departures from the behavior of a simple model
(the ideal gas) that holds under a limiting conditions (that of low density and therefore negligible
intermolecular interactions), the behavior of liquid solutions is discussed in terms of departures from
the model of ideal solutions, which holds in the limit of almost negligible differences in properties
between the solution components.
IDEAL SOLUTIONS The molecular picture of an ideal gas mixture is one with no intermolecular interactions. For a
condensed phase (solid or liquid), the molecules are closer each other; therefore unlike gas phase and
we could never assume no intermolecular interactions. Our molecular picture of an ideal solution will
be a solution where the molecules of the various species are so similar to one another that molecules
of one component can replace molecules of another component in the solution without changing the
solution’s spatial structure or the energy of intermolecular interactions.
Consider a solution of two species A and B. To prevent change in spatial structure of the liquids on
mixing A and B, the A molecules must be essentially the same size and shape as the B molecules. To
prevent change in the intermolecular interaction energy on mixing, the intermolecular interaction
energies should be essentially the same for A-A, A-B and B-B pairs of molecules.
EXTENSIVE AND INTENSIVE PROPERTIES An extensive property is the one that is equal to the sum of its values for the parts of the system. Thus,
it depends on the quantity. The mass of the system is the sum of the masses of the parts; mass is an
extensive property. So is volume. Properties that do not depend on the amount of matter in system are
called intensive. Ratio of the extensive property results in an intensive property. Density, molar
volume and pressure are examples of intensive properties.
PARTIAL MOLAR VOLUME Suppose we form a solution by mixing (at constant temperature and pressure) n1, n2,......nj moles of
substances 1,2,......,j. Let V1*,…. Vj* be the molar volumes of pure substances and let V* be the total
volume of the unmixed components at T and P.
V* = n1 V1*+ n2 V2*+…… nj Vj* (1)
After mixing, one finds that the volume V of the solution is not in general equal to the unmixed
volume; V≠ V1+V2. This is because intermolecular interactions in the solution differ from those in the
pure components. A similar situation holds for other extensive properties, for example, U, H, S, A, G.
Extensive properties of solutions (e.g. volume, energy, free energy etc.) are not truly additive when
pure components are mixed. Thus, when two liquids are mixed a decrease or an increase in the
volume may occur.
If F is any extensive property, then
F = f(P, T,n1,n2, .., nj) where nJ is the number of moles of the (J) th component of the system.
On differentiating,
25
Is called as “partial molar property of constituent,
Hence partial molar volume of a substance J at some general composition is
When the compositon of a two component mixture (A+B) is changed by the addition of dnA of A and
dnB of B, then the change in the total volume of the mixture is
dV= (∂V/∂nA) P,T,nB dnA + (∂V/∂nB)P,T,nA dnB = VA* dnA + VB dnB (3)
For a two component system of A and B, mean molar volume, Vm is
At constant pressure and temperature
26
Fig. 1. Determination of partial molar volume by intercept method
27
Thus, plotting the mean molar volume, Vm against the mole fraction xB, at the tangent at point O
corresponding to a definite xB and Vm is extrapolated to xB=0 (xA=1) intercept equals to molar volume of
A, VA* when the tangent at point O is extrapolated to xB=1 (xA=0) intercept equals to molar volume of B,
VB* as shown in Figure 1.
PURPOSE The aim of this experiment is to determine partial molar volumes of components of a binary mixture.
APPARATUS AND CHEMICALS Apparatus: Pycnometer, analytical balance.
Chemicals: Ethanol and water.
PROCEDURE i. Prepare 50 g of ethanol - water solutions with weight percentages 20, 40, 60 and 80%.
ii. For pure water, pure ethanol, 20, 40, 60 and 80 % ethanol - water mixtures perform the following.
1. Dry pycnometer and weigh it, mo.
2. Fill it with the solution.
3. Insert the plug with moderate pressure.
4. Weight the pycnometer again, mi.
TREATMENT OF EXPERIMENTAL DATA 1. Calculate weight of each solution and fill Table 1 in the report sheet.
2. Find volume of pycnometer, V0, using the density of water. (See your assistant for density of water
at the room temperature.)
3. Calculate number of moles of water and ethanol, nA,i and nB,i in each solution. (Pay attention the
percentage of ethanol used.)
4. Calculate mole fraction of ethanol, xB, in each solution.
5. Calculate the mole fraction of water, xA, in each solution.
6. Determine molar volume of pure water, VA* and ethanol, VB* using their densities.
7. Calculate mean molar volume, Vm, of each solution (using Eq. 4).
8. Plot Vm versus xB and determine the slope at each xB (dVm /dxB) and record in Table 2 in the report
sheet. (Graph should be polynomial with order of 5)
9. Plot Vm versus xA and determine the slope at each xA (dVm /dxA) and record in Table 2 in the report
sheet. (Graph should be polynomial with order of 5)
10. Determine partial molar volume of water and ethanol, VA* and VB* for each solution from the
graph.
Vm = VA* + xB (dVm/dxB)
Vm = VB* + xA (dVm/dxA)
QUESTIONS 1. What are the intensive and extensive properties of solutions. Why are the extensive properties
additive?
2. Discuss partial molar properties of solutions. Is this property intensive or extensive?
3. What are the ideal and real solutions?
4. Discuss the change in total volume as the % of ethanol in solution increases.
28
DATA SHEET
Fill the following table.
mi msi V0 nA,i nB,i XB XA
Pure water
20 % ethanol
40 % ethanol
60 % ethanol
80 % ethanol
___% ethanol
For pure water; VA* =
For ethanol; VB*
Vm (mL) (dVm /dxB) (dVm /dxA) VA* (mL) VB* (mL)
Pure water
20 % ethanol
40 % ethanol
60 % ethanol
80 % ethanol
___% ethanol
29
Unit–5
REDOX BEHAVIOR BY CYCLIC
VOLTAMMETRY AND DETERMINATION
OF ELECTROCHEMICAL PARAMETERS
30
CYCLIC VOLTAMMETRY
INTRODUCTION
BACKGROUND Cyclic voltammetry is the most widely used technique for acquiring qualitative information about
electrochemical reactions. This technique is based on varying the applied potential at a working electrode
in both forward and reverse directions (at some scan rate) while monitoring the current.
The equipment required to perform cyclic voltammetry consists of a conventional three-electrode
potentiostat connected to three electrodes (working, reference and auxiliary) immersed in a test solution.
The potentiostat applies and maintains the potential between the working and reference electrode while at
the same time measuring the current at the working electrode. Charge flows between the working
electrode and the auxiliary electrode. A recording device (such as a computer or plotter) is used to record
the resulting cyclic voltammogram as a graph of current versus potential.
IMPORTANT PARAMETERS IN CYCLIC VOLTAMMETRY The important parameters in a cyclic voltammogram are the peak potentials (Epc , Epa) and peak currents
(ipc , ipa) of the cathodic and anodic peaks, respectively. If the electron transfer process is fast compared
with other processes (such as diffusion), the reaction is said to be electrochemically reversible, and the
peak separation is
ΔEp = ׀Epa-Epc2.303 = ׀Rt/nF (1)
Thus, for a reversible redox reaction at 25 °C with n electrons ΔEp should be 0.0592/n V or about 60 mV
for one electron. In practice this value is difficult to attain because of such factors as cell resistance.
Irreversibility due to a slow electron transfer rate results in ΔEp > 0.0592/n V, greater, say, than 70 mV
for a one-electron reaction.
The location of the anodic (Epa) and cathodic (Epc) peaks on the potential axis are important parameters
which help us to calculate formal potential E°for a redox couple by averaging the two peak potentials.
E° = Epa + Epc/2 (2)
E° is characteristic of a redox species in much the same way that the wavelength of maximum absorbance
is characteristic of a species in spectroscopic experiments.
For a reversible reaction, the concentration is related to peak current by the Randles–Sevcik expression (at
25 °C):
ip = 2.686 x105 n
3/2 Ac° D
1/2 ν
½ (3)
whereip is the peak current in amps, A is the electrode area (cm2 ), D is the diffusion coefficient (cm
2 s
-1 ),
c° is the concentration (mol cm-3
), and n is the scan rate (V s-1
).
Peak currents will increase linearly as a function of the square root of the scan rate for reversible electron
transfer. Plots of ipvs. ν1/2
are useful in the characterization of electrochemically reversible redox systems.
Deviations from linearity are indicative either of complications in the kinetics of the observed electron
transfer, or the result of chemical changes which occur as a result of the electron transfer (homogeneous
reactions).
31
OBJECTIVE This unit will introduce you to cyclic voltammetry as a simple, rapid, and powerful method for
characterizing the electrochemical behavior of analytes that can be electrochemically oxidized or reduced.
EXPERIMENT NO. 1
STUDY OF FERRICYANIDE BY CYCLIC VOLTAMMETRY
PURPOSE In this experiment, the Fe(CN)6
-3/Fe(CN)6
-4 couple is used as an example of an electrochemically
reversible redox system . As the potential is scanned positively and is sufficiently positive to oxidize
Fe(CN)6-4
, the anodic current is due to the electrode process:
Fe(CN)6-4 Fe(CN)6
-3 e (4)
The electrode acts as an oxidant and the oxidation current increases to a peak. The concentration of
Fe(CN)6-4
at the electrode surface depletes and the current then decays. As the scan direction is switched
to negative, for the reverse scan the potential is still sufficiently positive to oxidize Fe(CN)6-4
, so anodic
current continues even though the potential is now scanning in the negative direction. When the electrode
becomes a sufficiently strong reductant, Fe(CN)6-3
, which has been forming adjacent to the electrode
surface, will be reduced by the electrode process:
Fe(CN)6-4
+ e Fe(CN)6-3
(5)
resulting in a cathodic current which peaks and then decays as Fe(CN)6-3
in the solution adjacent to the
electrode is consumed. In the forward scan Fe(CN)6-3
is electrochemically generated from Fe(CN)6-4
(anodic process) and in the reverse scan this Fe(CN)6-3
is reduced back to Fe(CN)6-4
(cathodic process).
Note that the technique of CV rapidly generates various oxidation states.
EQUIPMENT REQUIRED Potentiostat, small volume electrochemical cell, electrodes (glassy carbon working electrode, Pt auxiliary
electrode,and Ag/AgClreference electrode), Electrode polishing kit, volumetric flasks and micro pipette.
CHEMICALS REQUIRED 50 mL solution of 10 mM K3Fe(CN)6 in 1 M KCl.
PROCEDURE 1. Polish the electrode surface to a mirror finish with alumina slurry and rinse well with DI water and
dry it.
2. Fill the electrochemical cell with 10 mL of 10mM K3 Fe(CN)6 in 1 M KCl.
3. Degas the solution by bubbling Ar or N2 through it for ~30 sec and then move the bubbler up so
that it blankets the solution with gas but does not agitate it.
4. Set the potentiostat and make the electrode connections.
5. Run the CV between +0.60 V and -0.10 V, one full cycle, at 10mV/s.
6. Record CVs at 10, 20, 50, 100, 200, and 500mV/s.
TREATMENT OF RESULTS 1. Determine ipc,ipa, Epc, and Epa for each of the ferricyanide CV’s.
2. Make a plot of ipc vs. ν1/2
and ipc vs. ν1/2
for ferricyanide reduction and oxidation respectively.
3. From the slope of this plot and the area of your electrode, estimate the diffusion coefficient of
ferricyanide. The literature value is 7 x 10-6
cm 2/s at 25
o C.
32
4. Determine ΔEp = |Epa-Epc| at each scan rate. A reversible electrode process should have a peak
splitting of ~59/n mV.
EXPERIMENT NO. 2
CYCLIC VOLTAMMETRIC STUDY OF ASCORBIC ACID OXIDATION AT GLASSY
CARBON ELECTRODE
Purpose: Ascorbic acid (AA) or vitamin C is widely known as an antioxidant and a free radical scavenger. It is also
important in helping to produce collagen, a protein needed in the development and maintenance of bones,
cartilage, joint linings, skin, teeth, gums and blood vessels. AA is a water-soluble vitamin found mainly in
fruits and vegetables, particularly green leafy vegetables, citrus fruits, tomatoes, guavas, melons and
berries. Cyclic voltammetry (CV) is a convenient electrochemical method to examine the oxidation of AA
at a glassy carbon (GC) electrode.Electrochemical redox reaction of AA in different supporting
electrolytes media in both acidic and basic medium has been investigated in this experiment.
Equipment required: Potentiostat, small volume electrochemical cell, electrodes (glassy carbon working electrode, Pt auxiliary
electrode,and Ag/AgCl reference electrode), Electrode polishing kit, volumetric flasks and micro pipette.
Chemicals required: 50 mL solution of 10mM AA in Phosphate buffer solutions of pH2, 5, 7 and 9 respectively.
Procedure: 1. Clean the electrode surface prior to experiment. Polish the electrode (GC) with 1 micron or smaller
particles of alumina on a clean flat surface, preferably glass. Use of light pressure on the electrode
while tracing a figure 8 for 8- 10 times is suggested. Rinse the electrode with pure water and
sonicate (if available) for 1 minute in a beaker containing pure water. Wipe the edges of the
electrode (never the electrodes surface) with clean tissue paper and use immediately.
2. Fill the electrochemical cell with 10mL of phosphate buffer solution of desired pH containing
10mM AA.
3. Insert the working electrode (glassy carbon), the reference and the auxiliary electrode in the cell
andconnect theelectrodes to the potentiostat.
4. Select the potential window from 0.0 V to +0.6 V and carry out cyclic voltammetry at scan rates of
10mV/s, 20mV/s, 50mV/s 100mV/s and 200mV/s.
5. Repeat the steps for all solutions with varying pH.
Treatment of results: 1. Plot ip vs. ν
1/2 for the oxidation of AA from results obtained at pH 2, 5, 7 and 9.
2. Does the Ep value for the oxidation of AA shift with pH?
33
EXPERIMENT NO. 3 – PART 1
CYCLIC VOLTAMMETRY OF DINITROBENZENE AND AN ANTIOXIDANT
PURPOSE Antioxidants belong to a class of organic compounds that have an ability to inhibit or at least delay
oxidation processes occurring during cell metabolism and protect cells from the damage induced by
reactive species. They are present in a wide range of concentrations in body fluids and tissues by way of
synthetic processes in the body or by obtaining from the diet. Antioxidants are considered to play an
important role in biological systems by their anti-effects toward inflammations, viral, microbial and
fungal infections, cancer, tumor and cardiovascular diseases. This electrochemical experimentis based on
the interaction of an antioxidant such as Ascorbic acid with dinitrobenzene (DNB) which istoxic and
harmful to human health and environment and causes several diseases like neural disorder, anemia, liver
diseases and cancer.DNB, finds applications in industry as an industrial solvent (for perfumes and drugs)
and as chemical intermediate for dyes and pesticides.
EQUIPMENT REQUIRED Potentiostat, small volume electrochemical cell, electrodes (glassy carbon working electrode, Pt auxiliary
electrode,and Ag/AgCl reference electrode), Electrode polishing kit, volumetric flasks and micro pipette.
CHEMICALS REQUIRED Ascorbic acid, 1, 3-Dinitrobenzene, 10 mL Dimethyl sulfoxide (DMSO), Tetrabutylammonium tetra
fluoroborate (TBABF4).
PROCEDURE 1. Prepare 0.1M TBABF4 in 10mL of solvent (DMSO) and place in the electrochemical cell.
2. Purge the solution with Ar gas for 10-15 min. in order to remove any dissolved oxygen.
3. Take a blank CV run with the solvent (DMSO) containing 0.1 M TBABF4 in the potential range 0-
1.5V at 100mV/s scan rate.
4. Add appropriate amount of dinitrobenzene into the solution so that the concentration of
dinitrobenzene is 1mM and record CV.
5. Now add different amounts of the antioxidant (ascorbic acid) corresponding to final concentrations
of 0.5, 1, 2, 3, 4, 5 and 10 mM and record CV’s.
6. For comparison also record CV of ascorbic acid in DMSO in the potential range -2 - +1.5V at
100mV/s scan rate.
TREATMENT OF RESULTS 1. Evaluate electrochemical parameters such as peak potential (Ep), half-wave potential (E1/2), and
peak current (ip) – for the reduction of 1,3-DNB before and after the addition of various
concentrations of antioxidants.
2. Note the change in the oxidation peak current and finally irreversibility in 1, 3-DNB radical anion
and dianion systems upon the addition of antioxidant which reveals their interactions.
34
EXPERIMENT NO. 3 – PART 2
STUDY OF CHEMICALLY COUPLED COMPLEX REACTION OF
ANTIOXIDANT USING DINITROBENZENE RADICAL SYSTEM
PURPOSE This part of the experiment is an extension of the experimental procedures conducted in experiment 3(part
1) and provides further treatment of results.
Pure 1, 3-DNB shows two reversible reduction peaks in aprotic media. Two reversible peaks correspond
to one electron reduction processes. In cyclic voltammogram scanned for 1, 3-DNB system from 0 to
−1.5V, first reduction peak arrives due to the formation of the anion radical, 1,3-DNB•−
at Ecp= −0.800V
in DMSO, whereas the second reduction peak due to the formation of dianion, 1,3-DNB2−
is observed
around Ecp= −1.183V in DMSO. Both radical anion and dianion of 1, 3-DNB are stable species in aprotic
solvents.
Ascorbic acid is a primary antioxidant and can stop chain reaction by donating an electron to the free
radicals, and thus stop the propagation steps. The presence of 2-hydroxyl groups at the allylic (the most
labile) position is responsible for free radical scavenging activity. The present throws light on the
voltammetric behaviour of anion radicals and dianion of 1,3-DNB in the presence of antioxidant (ascorbic
acid). The addition of antioxidant to 1, 3-DNBs system and carrying out cyclic voltammetry to produce
anion radical and dianion shows marked effect on the voltammetric response as compared to that in the
absence of antioxidants. Progressive addition of antioxidants may lead to irreversibility of the anion
radical and dianion peaks in the voltammogram.
Anion radical and dianion are formed in a reversible electrochemical step conforming to pseudo first
order rate constants, i.e., kf, Eq. (1). Interaction between antioxidant and 1,3-DNB system involves
reversible electron transfer followed by irreversible chemical reaction mechanism (ErCi) as observed
through voltammetric responses:
1, 3-DNBe−
↔ 1, 3-DNB•−e−
↔1, 3-DNB2−
(1)
Second successive step involves the irreversible chemical reaction via protonation of 1,3-DNB anion
radical and dianion. These steps determine a second order rate constant, k2, Eqs. (2) and (3):
1, 3-DNB•−
+ AH → Products (2)
1, 3-DNB2−
+ AH → Products (3)
The possible mechanistic path may involve proton transfer and hydrogen atom abstraction from the
reactive moieties of ascorbic acid to anion radical and dianion. Intermediate reduction products along
with respective radicals of ascorbic acid may be formed via such interactions. The resultant antioxidant
radicals are electrochemically inactive in the working potential range.
TREATMENT OF RESULTS 1. Calculate pseudo first order rate constants (kf) for the anion radical and the dianion of 1,3-DNB in
the presence of ascorbic acid. Using Nicholson and Shain equation:
Ep = E1/2– RT/nF (0.78 – ln )
Where:a = RT/nF ν, E1/2 is the half-wave potential of 1,3-DNB anionradical and dianion. The values
of E1/2 arecalculated from the first and second reduction wavesin DMSO, respectively. Ep is the
peakpotential after the addition of higher concentration of ascorbic acid (10mM).
35
2. Calculate the second order rate constant, (k2), from therelation: k2 = kf/[AO]
Where: [AO] being the concentration of antioxidant which was present in large excess in order to
obtain pseudo first order conditions.
36
37
Unit–6
HYDROLYSIS OF ACID-CATALYZED ESTER
38
EXPERIMENT 1
A KINETIC STUDY OF THE HYDROLYSIS OF METHYL ACETATE
INTRODUCTION The hydrolysis of the ester methyl acetate in aqueous solution follows the stoichiometry
The reaction is catalyzed by hydrogen ions and follows the second order rate law
Rate = k [CH3CO2CH3] [H+] (1)
Because the hydrogen ion acts as a catalyst the reaction behaves as a first order reaction with the rate law
Rate = k' [CH3CO2CH3] where k' = k [H+]. (2)
This reaction can be followed analytically by sampling the reaction mixture and titrating the samples with
sodium hydroxide for the total acid concentration. The acid in the reaction mixture consists of H+ from
the catalyst (HCl) and the acetic acid, CH3CO2H, produced by the reaction. The total acid concentration
and the HCl concentration allow one to calculate the CH3CO2H concentration.
One experiment was carried out to determine values for both the apparent first order rate constant k' and
the second order rate constant k.
APPARATUS REQUIRED Five 50 mL and two 100 mL conical flasks with corks, beakers, burette, pipettes of 25mL, 10 mL, 5mL
and 2 mL, a thermostat or water bath, stopwatch.
CHEMICALS REQUIRED 0.5 N HCl, 0.05 N Sodium hydroxide, methylacetate, ice cold distilled water, phenolphthalein.
PROCEDURE 1. Take 50mL of 0.5N HCl in a clean 100 mL conical flask and about 10mL pure methyl acetate in a
test tube, cork both of them and place them in a thermostat maintained at room temperature.
2. Maintain a stock of about 200mL of ice cold CO2 free water.
3. Arrange five conical flasks of 50 mL capacity and put 25mL of ice cold water in each flask in turn
before the titrations of aliquots of the reaction mixture during the course of the reaction.
4. When both the acid and the ester attain constant temperature, measure out methylacetate with a 2
mL pipette from the test tube into the conical flask containing 50 mL of hydrochloric acid solution.
5. Start the stop-watch when half the pipette has been discharged. Shake the mixture at once take out
2mL of the reaction mixture and transfer it to one of the flasks containing 25mL of ice cold water.
This freezes or arrests the reaction.
6. Titrate it quickly against 0.05N NaOH using phenolphthalein as indicator and find out the volume
of the alkali required for the titration. This volume of the alkali corresponds to the concentration of
HCl in the reaction mixture before the hydrolysis starts.
39
7. Bulk of the expected alkali volume should be added quickly in each succeeding titration and then
the titration should be completed carefully with shaking, till the first appearance of pink color
which does not vanish in 10 seconds.
8. 2 mL of the reaction mixture are taken out at successive intervals of 10, 20, 30, 40, and 50 minutes
and are in turn transferred to flasks containing 25mL of ice cold water and titrated against standard
alkali.
9. The volume of alkali required for each titration is noted.
10. Note the infinite time reading i.e. the volume required when the hydrolysis is complete. Take out 10
mL of the reaction mixture in a dry small conical flask and loosely cork it. Put this corked flask in a
water bath maintained at 50oC for an hour. Cool it to room temperature. Take out 2 mL of this
mixture and titrate as before against 0.05N alkali. This titre value corresponds to the total acid
concentration after hydrolysis is complete.
RESULTS AND DISCUSSION The titration volume used to titrate the acetic acid product was calculated using the equation.
Which in turn was used to calculate the concentration of unreacted methyl acetate according to the
equation
[CH3CO2CH3] = 0.100 M – [CH3CO2H] (4)
The apparent first order rate constant for the reaction was determined using the integrated first order rate
equation
ln [CH3CO2CH3] = –k't + ln [CH3CO2CH3] (5)
from which it can be seen that a plot of ln [CH3CO2CH3] versus time should yield a straight line with
slope equal to –k'.
The data from the experiment along with the transformed data for determining the apparent first order rate
constant k' are found in table 1.
Table 1
Data and Calculations for Determining First Order Rate Constant for the Hydrolysis of Methyl Acetate
Time (min) Titration Titre Volume [Acetic Acid] [Methyl Acetate] ln [Methyl
Volume (mL) for Acetic Acid (mL) (M) (M) Acetate]
1.25 26.12 1.12 0.0045 0.0955 -2.348
10.20 32.45 7.45 0.0298 0.0702 -2.656
19.55 37.35 12.35 0.0494 0.0506 -2.984
31.20 41.67 16.67 0.0667 0.0333 -3.402
40.75 43.93 18.93 0.0757 0.0243 -3.718
50.50 45.75 20.75 0.0830 0.0170 -4.075
These data were used to construct a plot of ln [Methyl Acetate] versus time (see Figure 1) which
according to equation (6) has a slope equal to –k'. The slope of the best straight line through the data was
calculated using linear regression to be –0.0350 min-1 with correlation coefficient of –0.9999. This means
Titre Volume for CH3CO2H = Total Titre Volume – 25.0 mL. (3)
40
that the apparent first order rate constant k' for the reaction is 0.0350 min-1 from which the second order
rate constant k is calculated to be 0.350 L mol-1 min-1 using equation (2).
In summary, the hydrolysis of methyl acetate was found to behave as a first order reaction in 0.100 M
HCl with an apparent first order rate constant k' of 0.0350 min-1
and a second order rate constant k of
0.350 L mol-1
min-1
at 25°C.
CALCULATIONS
[CH3CO2CH3] = .100 M – 0.00450 M
= .0955
ln[CH3CO2CH3] = –2.348
Slope for plot of ln[CH3CO2CH3] vs time is 0.0350 min–1
Therefore k' = 0.0350 min–1
Example calculation for Sample #1 Time 1.25 minutes, [H
+] = 0.100 M
Titre Volume for CH3CO2H = 26.12 – 25.0 mL
= 1.12 mL
[CH3CO2H] =
0.100 M x 1.12 mL
25.0 mL
= 0.00450 M
41
EXPERIMENT 2
THE HYDROLYSIS OF ETHYL ACETATE
OBJECTIVES Determine the value of the equilibrium constant for a reaction
Use acid-base titrations and solution stoichiometry in determining the equilibrium constant
BACKGROUND MATERIAL A state of balance in a chemical reaction in which the forward and backward rates are equal is equilibrium
For the reaction below, when A and B were mixed, the reaction proceeds in the forward direction to
produce C and D. However, as time progresses, the concentration of C and D increases causing an
increase in the rate of the reverse reaction. Concurrent with this increased rate of the reverse reaction is a
reduction of the forward rate due to the decrease in the concentration of A and B. At some point, the rate
of the forward and reverse reactions will become the same and we will reach a state of dynamic
equilibrium:
aA + bB cC + dD
Equilibrium does not mean that the forward and reverse reactions have stopped. Molecules of A and B
are still reacting to form C and D and molecules of C and D are reacting to form A and B. However,
since the rate of the forward and reverse reactions is the same, it will appear that nothing is happening.
As such, all quantifiable physical and chemical properties such has pH, color, and concentration will
remain constant.
For a general equilibrium equation we can specify an equilibrium constant, Kequil, that relates the
concentrations of all product and reactant species,
where [A], [B], [C], and [D] are the molar concentration of all species present at equilibrium. The
exponents, a, b, c, and d represent the stoichiometry coefficients from the balance chemical reaction.
Kequil is really the ratio of the rate of the reverse reaction divided by the rate of the forward reaction and so
is a dimensionless constant at a given temperature.
We will be studying the acid catalyzed (HCl) hydrolysis of an ester (ethyl acetate, EtOAc), to form an
alcohol (ethanol, EtOH) and an acid (acetic acid, HAc):
CH3CH2CO2CH3 + H2O CH3CH2OH + CH3CO2H
EtAc + H2O EtOH + Hac
This hydrolysis reaction will occur spontaneously at room temperature and the addition of acid only
affects the amount of time it takes the reaction to reach equilibrium.
In practice it is often difficult to determine the equilibrium concentration of all these species. However,
we can calculate the equilibrium constant by determining the concentration of a single species at
equilibrium, if we know the initial concentration of all species. We do not have to be concerned about the
concentration of the hydrochloric acid since it only serves as a catalyst. For our specific reaction, the
equilibrium constant will equal:
42
Since we will know the initial concentration of each component, we only need to determine the
equilibrium concentration of one component to calculate the equilibrium constant. Although we have
several chemical species to choose from, we will monitor the acetic acid concentration since it can be
very accurately determined via a simple titration with standardized base.
In the above discussion, we decided to monitor the concentration of the HAc to determine the equilibrium
constant. However, the presence of the strong acid HCl complicates our task. A simple titration with
sodium hydroxide will neutralize all the acids that are present and we will not be able to differentiate the
amount of base needed to neutralize the HCl versus the amount that was needed to neutralize the HAc.
This is a common problem in chemistry and is solved by titrating a blank, which only contains the HCl
(Solution 1). Since it contains a known amount of HCl and water, a simple titration will determine the
amount of base needed to neutralize the HCl contained in the other solutions. This amount of HCl will be
constant in the other solutions because the HCl is a catalyst that is not consumed by the reaction. Now
that we know the amount of base needed to account for the HCl, any additional base used in the titration
of Solution #3/4, Solution #5/6, and Solution #7/8, must represent the amount of HAc present.
SAMPLE CALCULATION The volume of 1.090 M NaOH used for titrations was:
Vials 1 & 2 Vials 3 & 4 Vials 5 & 6 Vials 7 & 8
13.48 ml 34.74 ml 20.25 ml 29.60 ml
VIALS 1 & 2 Remember that molarity x volume = moles so:
1.090 M x .01348 L = .0147moles HCl
To get grams of HCl present:
.0147 moles HCl x 36.45 g/mole = 0.536 g HCl in 5 mL
What is the total mass of the 5 mL of HCl?
1.0436 g/mL x 5 mL = 5.218 g HCl
What is the mass of water in the 5 mL of HCl?
5.218 g - 0.536 g = 4.682 g H2O
How many moles of water present in the 5 mL of HCl?
4.682 g H2O / 18 g/mole = 0.260 mole water in 5 mL
VIALS 3 AND 4 It may help you to set up a small table looking at the initial and final conditions of the reaction
EtOAc H2O EtOH HAc
Initial: 3 mL 2 mL + 0.260 mol 0 mL 0 mL
Change: - x - x + x + x
You get x from the titration information:
Moles base used in titration:
1.090 M NaOH x .03474 L = 0.03787 moles base
43
Since the stoichiometry of the reaction is 1 to 1, moles base = moles acid:
0.03787 moles base = 0.03787 moles acid
This is NOT x because the titration only gives you moles of total acid, dont forget about HCl:
0.03787 moles acid 0.0147 moles HCl = 0.02317 moles acetic acid = x
Final moles ethanol is x to so moles ethanol = 0.02317 moles = x
How many moles of ethyl acetate present initially:
3 mL x 0.894 g/ml 88g/mol = 0.03056 moles ethyl acetate initially
How many moles of water initially:
2 mL x 1 g/mL 18 g/mol = 0.1111 moles water initially
Lets revisit the table:
EtOAc(aq) H2O(soln) EtOH(aq) HAc(aq)
Initial: .03056 mole 0.1111 + .026 mole 0 mole 0 mole
Change: -.02317 mole -.02317 mole +.02317 mole +.02317 mole
Equilibrium: .00739 .34793 mole .02317 mole .02317 mole
Now we can calculate the equilibrium constant for this reaction as:
Kequil = Kequil = 0
Repeat the above. Be careful with x!!!. Remember that the titration only tells you the moles of total acid
present. Vials 5 and 6 already have acid present initially.Watch out!!!
VIALS 7 & 8 Repeat the above.
44
EXPERIMENT 3
KINETICS OF FIRST ORDER REACTIONS
OBJECTIVE
To follow the reaction by a titrimetric method and determination of the first order rate constant by a
graphical.treatment.of.the.data.
INTRODUCTION When an ester (such as ethyl acetate) is mixed with water it is converted into alcohol and acid according
to the following equation:
When the amount of water is relatively large, the reaction goes practically to completion (the equilibrium
shifts to the right) and the rate is first order with respect to the ester. The hydrolysis takes place slowly
with pure water and is catalyzed by acids.
The elementary step in a first order reaction is represented by the following equation
k1
A (reactant) ---------> P (product) (2)
Where k1 is the first order rate constant.
By definition, the rate of a first order reaction is equal to the rate of loss of product or the rate of
formation of product and is proportional to the concentration of reactant,
i.e. -d[A]/dt = +d[P]/dt α [A] (3)
or -d[A]/dt = k1 [A] (4)
Rearranging equation 4 and integrating gives
ln[A] = -k t + const (5)
if at time t = 0 and time t = t, the concentration of A is [A]o and [A]t respectively, the constant of
integration is found to be ln[A]o, and equation 5 becomes
ln[A]t = -k1t + ln[A] (6)
and a plot of ln[A] against 't' should give a slope of -k1.
MATERIALS 25 cm
3 and 5 cm
3 pipettes, 0.5 M HCl, 0.1 M NaOH, ethyl acetate, crushed ice, potassium hydrogen
phthalate, stop watch.
EXPERIMENTAL
1. Pipette 100.0 cm3 of 0.5 M HCl into a conical flask (labeled A) and a further 20.0 cm
3 into a second
conical flask (labeled B).
2. Prepare 25 cm3 of crushed-ice/water. Pipette 5.0 cm
3 of ethyl acetate into flask A, shake well, start
the stop watch and immediately withdraw 5.0 cm3 of the solution. Immediately, run this into the
crushed-ice water mixture, and swirl to 'stop' the reaction, and as soon as possible titrate with the
NaOH solution. Without ever stopping the watch, note the time to the nearest second at which the
solution is run into the ice/water mixture. This is the first time, t1, of the series.
45
3. About 10 min after the first withdrawal pipette another 5.0 cm3 of the reaction mixture from the
flask and titrate as before, again noting the exact time, t2, that the reaction is 'stopped'.
4. Further titrations are made at times of about 20, 30, 40, 60, 80, 100 and 120 min.
5. The reaction is accelerated to completion by heating to 70°C for ½ hr. During this time the flask
must be stoppered to avoid changes in concentration due to evaporation. Cool to room temperature,
pipette 5.0 cm3 and add to 25 cm
3 of water and titrate as before.
6. For a control, take flask B and add 1 cm3 of water, pipette 5.0 cm
3 and titrate as before.
CALCULATION
The initial ethyl acetate concentration is proportional to (V∞ - Vo), where V∞ is the final titre volume for
flask A and Vo is the titre volume for flask B. The concentration of ethyl acetate at time t is proportional
to (V∞ - Vt) where Vt is the titre volume of the sample at time t. Plot ln(V∞ - Vt) against time. Hence
determine the value of k1 (in s-1
) and the half-life of the reaction.
EXERCISE
1. If the temperature of the reaction increased by about 2°C during the course of the reaction, what
effect would this have on the points plotted?
2. Assume the activation energy, Ea, to be about 11 kJ mol-1
, calculate the fractional variation in the
rate constant that would result from this increase in temperature.
3. What is the significance of the titration value obtained from flask B.
4. Would you expect the first titration value for flask A to be zero? Why?
5. Find the error in the slope of the graph using the “box” method described in the lab talk, and so
determine the uncertainty in k1.(See Appendix 6)
46
EXPERIMENT 4
KINETICS OF THE HYDROLYSIS OF ETHYL ACETATE
THEORY
Chemical kinetics, Chemical reactions, reaction rate
Chemical kinetics is the part of physical chemistry that studies reaction rates. The reaction rate or rate of
reaction for a reactant or product in a particular reaction is intuitively defined as how fast a reaction takes
place. For example, the oxidation of iron under the atmosphere is a slow reaction which can take many
years, but the combustion of butane in a fire is a reaction that takes place in fractions of a second.
Consider a typical chemical reaction:
aA + bB → pP + qQ A
The lowercase letters (a, b, p and q) represent stoichiometric coefficients, while the capital letters
represent the reactants (A and B) and the products (P and Q).
According to IUPAC's Gold Book definition the reaction rate (v) for a chemical reaction occurring in a
closed system under constant-volume conditions, without a build-up of reaction intermediates, is defined
as:
v
1 dc
A
1 dc
B
1 dc
P
1 dcQ
(1)
a
dt b dt p dt q dt
where: cI (I= A, B, P. or Q) – is concentration of substance
The IUPAC recommends that the unit of time should be the second always. Reaction rate usually has the
units of mol dm-3
s -1
. It is important to on mind that the previous definition is only valid for a single
reaction, in a closed system of constant volume.
The quantity:
d
(2)
dt
defined by the equation:
1 dn
A
1 dn
B
1 dn
P
1 dnQ
(3)
a
dt b dt p dt q dt
47
where: nI - designates the amount of substance I (I=A, B, P, or Q) conventionally expressed in units of
mole ξ is called the 'rate of conversion' (extent of reaction) and is appropriate when the use of
concentrations is inconvenient, e.g. under conditions of varying volume. In a system of constant volume,
the rate of reaction is equal to the rate of conversion per unit volume throughout the reaction.
The rate law or rate equation for a chemical reaction is an equation which links the reaction rate with
concentrations or pressures of reactants and constant parameters (normally rate coefficients and partial
reaction orders). To determine the rate equation for a particular system one combines the reaction rate
with a mass balance for the system. For a generic reaction:
A + B → C
the simple rate equation is of the form: B
v k cAacB
b (4)
the concentration is usually in mol dm-3
and k is the reaction rate coefficient or rate constant. Although
it is not really a constant, because it includes everything that affects reaction rate outside concentration:
mainly temperature, ionic strength, surface area of the adsorbent or light irradiation.
The exponents a and b are called reaction orders and depend on the reaction mechanism. The
stoichiometric coefficients and reaction orders are very often equal, but only in one step reactions,
molecularity (number of molecules or atoms actually colliding), stoichiometry and reaction order must be
the same.
The Arrhenius equation is a simple, but remarkably accurate formula for the temperature dependence of
the rate constant, and therefore rate of a chemical reaction. Actually, the Arrhenius equation gives:
"the dependence of the rate constant (k) of chemical reactions on the temperature (T) (in Kelvin) and
activation energy (Ea) ", as shown below:
where: A – is the pre-exponential factor or simply the prefactor R – is the molar gas constant.
The units of the pre-exponential factor are identical to those of the rate constant and will vary depending
on the order of the reaction. It can be seen, that either increasing the temperature or decreasing the
activation energy (for example through the use of catalysts) will result in an increase in rate of reaction.
The activation energy can be interpreted as the minimal energy of the molecules to undergo reaction.
This energy is needed, either, to rupture a chemical bond, eg. in free radical gas reactions, or to allow
rearrangements when the molecules collide.
Taking the natural logarithm of the Arrhenius equation yields:
ln k
Ea 1
ln A (6)
R T
So, when a reaction has a rate constant which obeys the Arrhenius equation, a plot of
1 ln k = f T
gives a straight line, which slope and intercept can be used to determine thr Ea and A.
48
This procedure has become common in experimental chemical kinetics. To determine the activation
energy of a reaction, one must know a rate constant of the reaction at least at two different temperatures.
Applying the Equation 6, one can easy express:
where: k1 – is rate constant correspond to temperature T1 k2 – is rate constant correspond to temperature
T2
Since the rate of a given reaction depends on the concentration of the reactants, the speed of the process
falls off as the reaction proceeds, for the reactants being continuously consumed. The reaction is
becoming slower and slower but theoretically never ceases. It is, therefore, not possible to define the
general rate of a reaction, and so in practice the rate is considered at a particular instant. The rate may be
defined in any convenient way, usually, the rate of change of concentration (c) of one of the reactants or
products is chosen. The experimental data then follow a change of concentration with time (t), and the
rate at any instant is given by the tangent to a curve of the plot:
c = f(t)
THE SECOND-ORDER REACTION depends on the concentrations of one second-order reactant (scheme C and Equqtion 8), or two first-
order reactants (scheme D and Equation 9):
2A → Products C
A + B → Products D
For a second order reaction, its reaction rate is given by:
v k c2
(8)
A
v kcA cB (9)
We will deal with the bimolecular reaction – D, supposing the same initial concentration of A and B
reactants:
c0 A
c0B
c0
The differential rate law for the second-order reaction is then:
dc
kc2
(10)
dt
Solving the differential equation, one can obtain:
49
where: c – is the concentration of reactant at time t → ( cA cB c )
k – is the second-order constant, which has dimension concentration-1
time-1
(eg.
dm3 mol
-1s
-1
In this case, a characteristic plot which will produce a linear function is 1/c /c f (t) with the
slope = k (Figure 1).
The half-life of reaction describes the time needed for half of the reactant to be depleted. The half-life of
a second-order reaction, which depends on one second-order reactant, is:
Figure 1 Plots c = f(t) and 1/c = f(t) for a second-order reaction
Task Determine the rate constant and the activation energy of the alkaline hydrolysis of ethyl acetate using
sodium hydroxide.
This experiment illustrates a bimolecular reaction (reacting species are ethyl acetate and sodium
hydroxide):
CH3 –COO–CH2–CH3 + NaOH → CH3–COONa + CH3–CH2–OH E
The initial concentrations of the reacting species are the same:
c0 A
c0B
where: c0 A - is the initial concentration of ethyl acetate
c0 B - is the initial concentration of sodium hydroxide
50
Equipments and chemicals Thermostat
Pipettes
Burette
Volumetric flasks, titrimetric
flasks stop-clock
solution of ethyl acetate (c = 0.04 mol dm-3
)
solution of sodium hydroxide (c = 0,04 mol.dm-3
)
solution of hydrochloric acid (c = 0,04 mol.dm-3
)
phenolphthalein
Procedure 1. Transfer 50 ml of the solution of ethyl acetate (c = 0.04 mol dm
-3) into a volumetric flask (V=50
ml) and 50 ml of the solution of sodium hydroxide (c = 0.04 mol dm-3
) into another volumetric flask
(V=200 ml).
Both flasks cork down and put them in the thermostated bath (t = 20 °C).
2. Fill the burette with the solution of sodium hydroxide (c = 0.04 mol dm-3
).
3. Pipette VHCl= 5 ml solution of hydrochloric acid (c = 0.04 mol dm-3
) into a clean and dry titrimetric
flask.
4. After 10 minutes, take out the flasks with the solutions from the thermostated bath and pour the
solution of ethyl acetate to the solution of sodium hydroxide, put the mixture to the thermostated bath.
Start the stop-clock.
5. 5 minutes after mixing, pipette 10 ml of reaction mixture (leave the flask in the bath!) to the
titrimetric flask (with 5 ml of HC - VHCl ).
Remark: HCl stops the reaction given by the scheme D.
6. Transfer 50 ml of the solution of ethyl acetate (c = 0.04 mol dm-3
) into a volumetric flask (V=50
ml) and 50 ml of the solution of sodium hydroxide (c = 0.04 mol dm-3
) into another volumetric flask
(V=200 ml).
Both flasks cork down and put them in the thermostated bath (t = 20 °C).
7. Fill the burette with the solution of sodium hydroxide (c = 0.04 mol dm-3
).
8. Pipette VHCl= 5 ml solution of hydrochloric acid (c = 0.04 mol dm-3
) into a clean and dry titrimetric
flask.
9. After 10 minutes, take out the flasks with the solutions from the thermostated bath and pour the
solution of ethyl acetate to the solution of sodium hydroxide, put the mixture to the thermostated bath.
Start the stop-clock.
10. Transfer 50 ml of the solution of ethyl acetate (c = 0.04 mol dm-3
) into a volumetric flask
V=50 ml) and 50 ml of the solution of sodium hydroxide (c = 0.04 mol dm-3
) into another volumetric
flask (V=200 ml).
Both flasks cork down and put them in the thermostated bath (t = 20 °C).
11. Fill the burette with the solution of sodium hydroxide (c = 0.04 mol dm-3
).
12. Transfer 50 ml of the solution of ethyl acetate (c = 0.04 mol dm-3
) into a volumetric flask (V=50
ml) and 50 ml of the solution of sodium hydroxide (c = 0.04 mol dm-3
) into another volumetric
flask (V=200 ml).
Both flasks cork down and put them in the thermostated bath (t = 20 °C).
13. Transfer 50 ml of the solution of ethyl acetate (c = 0.04 mol dm-3
) into a volumetric flask (V=50
ml) and 50 ml of the solution of sodium hydroxide (c = 0.04 mol dm-3
) into another volumetric
flask (V=200 ml).
Both flasks cork down and put them in the thermostated bath (t = 20 °C).
14. Fill the burette with the solution of sodium hydroxide (c = 0.04 mol dm-3
).
51
15. Pipette VHCl= 5 ml solution of hydrochloric acid (c = 0.04 mol dm-3
) into a clean and dry titrimetric
flask.
16. After 10 minutes, take out the flasks with the solutions from the thermostated bath and pour the
solution of ethyl acetate to the solution of sodium hydroxide, put the mixture to the thermostated
bath.
17. Start the stop-clock.
18. 5 minutes after mixing, pipette 10 ml of reaction mixture (leave the flask in the bath!) to the
titrimetric flask (with 5 ml of HC - VHCl ).
19. Remark: HCl stops the reaction given by the scheme D.
20. Titrate with the solution of sodium hydroxide adding 1 drop of phenolphthalein as indicator.
21. When the endpoint of titration has been reached, read the used volume of NaOH from the burette
(VNaOH). Write it down to the Table 1.
22. Repeat the step 5 and 6 every 5 minutes six times more (in the 10th, 15th, 20th, 25th, 30th and 35th
minutes from the moment of mixing).
23. Write down to the Table 1 the temperature of the bath.
24. Repeat the same experiment at 30 °C. Because the reaction is faster, the times for titrations will be
in the 5th, 10th, 15th, 20th, and 25th min. from the mixing.
Transfer 50 ml of the solution of ethyl acetate (c = 0.04 mol dm-3
) into a volumetric flask (V=50 ml)
and 50 ml of the solution of sodium hydroxide (c = 0.04 mol dm-3
) into another volumetric flask
(V=200 ml).
Both flasks cork down and put them in the thermostated bath (t = 20 °C).
25. Fill the burette with the solution of sodium hydroxide (c = 0.04 mol dm-3
).
26. Pipette VHCl= 5 ml solution of hydrochloric acid (c = 0.04 mol dm-3
) into a clean and dry titrimetric
flask.
27. After 10 minutes, take out the flasks with the solutions from the thermostated bath and pour the
solution of ethyl acetate to the solution of sodium hydroxide, put the mixture to the thermostated
bath.
Start the stop-clock.
28. 5 minutes after mixing, pipette 10 ml of reaction mixture (leave the flask in the bath!) to the
titrimetric flask (with 5 ml of HC - VHCl ).
Remark: HCl stops the reaction given by the scheme D.
29. Titrate with the solution of sodium hydroxide adding 1 drop of phenolphthalein as indicator.
When the endpoint of titration has been reached, read the used volume of NaOH from the burette
(VNaOH). Write it down to the Table 1.
30. Repeat the step 5 and 6 every 5 minutes six times more (in the 10th, 15th, 20th, 25th, 30th and 35th
minutes from the moment of mixing).
31. Write down to the Table 1 the temperature of the bath.
32. Repeat the same experiment at 30 °C. Because the reaction is faster, the times for titrations will be
in the 5th, 10th, 15th, 20th, and 25th min. from the mixing.
52
Write down to the Table1 the used volume of NaOH for each titration.
Table 1 Measured and calculated values
t = 20°C t = 30°C
t [min] VNaOH c 1/ c VNaOH c 1/ c [ml] [mol dm-3] [dm3 mol-1] [ml] [mol dm-3] [dm3 mol-1]
5
10
15
20
25
30 -
35 -
Data treatment 1 Calculate the concentration (c) in the Table 1 according to:
c
VHCl
cHCL
VNaOH
cNaOH
(13)
V
= 0.04 mol dm-3
) where: cHCl - is concentration of HCl (cHCl
5
VHCl- is the volume of HCl (5 ml)
cNaOH - is concentration of NaOH (cNaOH = 0.04 mol dm-3
)
VNaOH - is the volume of NaOH from Table 1 (ml)
V- is the volume of the reaction mixture used in titration (10 ml)
VHCl- is the volume of HCl (5 ml)
cNaOH - is concentration of NaOH (cNaOH = 0.04 mol dm-3
)
VNaOH - is the volume of NaOH from Table 1 (ml)
V- is the volume of the reaction mixture used in titration (10 ml)
1. Calculate the 1/c values.
2. Use MS Excell to create the dependence 1/c = f (t) at given temperature.
3. Fit the experimental points with a linear function. The slope represents the value of the rate constant
(k) at given temperature.
4. If the time is in minutes, the unit of the rate constants is dm3 mol
-1 min
-1. 4 Calculate the activation
energy (Ea) according to Equation 7.
Report The report must include:
Theoretical principles
Equipments and chemicals
Experimental procedure and measurements
Table of results, calculations, diagrams 1/c = f (t) at two temperatures, and the value of the activation
energy.
53
Unit–7
DETERMINATION OF RATE
CONSTANT FOR THE IODINATION OF
CYCLOHEXENE ABSORPTOMETRICALL
54
THE IODINATION OF CYCLOHEXENONE
Introduction In this experiment the kinetics of the reaction between triiodide and cyclohexanone (C6H10O)
will be studied.
Although we expect the rate of this reaction to be dependent upon the concentrations of cyclohexanone
and triiodide present in the solution, in somewhat of a surprise it is also dependent upon the hydrogen
ion concentration. The general rate law for this reaction is:
where a, b and c (generally integers) are the orders of the reaction with respect to cyclohexanone,
triiodide, and hydrogen ion, respectively, and k is the rate constant for the reaction.
For a particular set of conditions, the experimental of this reaction can be expressed as the change in the
concentration of I3−, Δ[I3−], divided by the time interval, Δt,
The average rate or velocity of a car to a given location is the distance traveled divided by time. The
amount of I3 − consumed is analogous to the distance traveled by the car; both quantities, when divided
by elapsed time, give rates. The rate is generally not constant, but varies with time because the
concentrations of the reactants decrease as they are consumed during the reaction. This complication
normally makes the study of reaction rates rather complex. However, in the reaction between triiodide
and cyclohexanone, the triiodide reacts or is "consumed", after the rate determining step. This permits us
to relate the rate of reaction to the disappearance of triiodide. Triiodide is a brown colored liquid, and the
reaction is considered to be over when the brown color just disappears.
If a series of reactions is run in which the concentration of only one of the reactants is changed at a time,
then the order of reaction with respect to that reactant can be determined. As an example, the order with
respect to cyclohexanone can be determined by examining the rates of two different trials. In both trials
the triiodide and hydrogen ion concentrations are held constant and only the cyclohexanone concentration
is varied. Thus, any change in the rate is due to the change in the cyclohexanone concentration. If one
now compares the rate of trial II with trial I the following expression results:
55
Since the triiodide and hydrogen ion concentrations are being held constant the equation reduces to:
By taking logs of both sides of the equation the order of reaction with respect to cyclohexanone can be
determined by:
rearranging
A similar procedure enables one to measure theorder of the reaction with respect to hydrogen ion and to
confirm the fact that the reactions order is zero with respect to triiodide. Once the order of reaction with
respect to each reactant is known, the rate constant, k, for the reaction can be evaluated.
In the second part of this experiment the energy of activation, Ea, will be evaluated by studying the rate of
the reaction at two additional temperatures. The procedure will be to inspect the rate of reaction of the
fastest trial mixture at a lower temperature and the slowest trial mixture at a higher temperature. Using
these rates and calculating the rate constants at each of the three temperatures permits one to estimate the
energy of activation. The relationship between the rate constant, temperature, and the energy of activation
is given by the Arrhenius Equation:
To find the energy of activation, ln k is plotted against 1/T. The slope is equal to -Ea/R. R is the gas
constant, 8.314 J/mol-1
K.
Experimental
Part I. Determination of Reaction Orders. 1. Select two 18 x 150 mm test tubes and fill them about three-quarters full with distilled water.
Both tubes must be filled to the same height. Hold the two tubes beside each other, place a sheet of
white paper under them and look down through the two tubes. If the colors appear the same in both
tubes then the pair can be used to observe the color change due to the loss of the I3−. If the colors
are different then select another tube and re-do the comparison test. These tubes will be used for all
of the following trials.
56
2. In a clean, dry Erlenmeyer flasks obtain approximately 75 mL of 0.50 M C6H10O (cyclohexanone),
and 75 mL of 1.0 M HCl. In a third flask obtain 50 mL of 0.020 M I3 −. Keep the flasks containing
the I3 − and C6H10O covered to prevent evaporation of the cyclohexanone and triiodide.
3. Preparation of a blank: Using graduated pipettes, transfer 6 mL of the 0.50 M C6H10O solution, 6
mL of the 1.0 M HCl solution, and 5 mL of distilled water into an Erlenmeyer flask. Separate
pipettes should be used with each solution. Lastly pipette 3 mL of the 0.020 M I3 − into the flask,
and swirl it to mix the reagents.
Let this mixture stand for 10 minutes and then fill one of the test tubes from A-1. This test tube will
be used as a comparison to determine when each of the of the following trials has fully reacted.
4. Use the graduated pipettes to transfer 6 mL of the 0.50 M C6H10O solution, 6 mL of the 1.0 M HCl
solution, and 10 mL of distilled water into an Erlenmeyer flask. Lastly pipette 3 mL of the 0.020 M
I3 − into the flask, and swirl it to mix the reagents. Since the reaction begins the instant the triiodide
is added it is important to record the time that the reagents are mixed to the second. Transfer this
solution to the second test tube until it is at the same height as the solution in the test tube being
used as the blank. The brownish color of the solution is due to I3 − . As the I3 − is consumed by the
reaction the color will fade. Keep looking down the two test tubes toward a piece of white paper
and note the time at which the colors in the tubes are the same. Do a second trial. The time required
for the two trials should agree within 10%. If the difference in the time for the two trials is greater
than 10%, a third trial should be performed.
5. If the concentration of one of the other components is changed while keeping the concentration of
the other two components constant, the dependence of the reaction upon that component can be
determined. Start by doubling the volume of C6H10O used. The amount of distilled water used will
have to be decreased, as the total volume of the reaction mixture should always be 25 mL.
Remember the [H+] and [I3 −] must be the same as in the initial trials. Only one reactant
concentration should be changed in each new mixture tested. Perform two trials at the new
cyclohexanone concentration. Do a third trial if the rates vary by more than 10%.
6. In the next series of trials double the [H+] , while returning the C6H10O concentration to that of the
original mixture. Do two trials at this H+ concentration. Do a third trial if the rates vary by more
than 10%.
7. Finally, reduce or increase the [I3 −] so that the time it takes for the triiodide to disappear is
reasonable. The [H+] and the [C6H10O] should be the same as the original mixture. Perform two
trials on this mixture. Do a third trial if the rates vary by more than 10%.
8. Record the temperature of the mixture from one of the trials. As all the bottles have been stored in
the same room it safe to assume that this is the temperature of all the solutions.
Energy of Activation The energy of activation, Ea, can be determined by measuring the rate constant, k, at two different
temperatures and then plotting the ln k against l/T.
1. Select the reaction mixture that took the longest time and measure the rate of this reaction in the
hotter water bath, at about 33°C. Be sure to use the solutions of C6H10O, H+, I3 − and distilled
water that are suspended in the water bath to perform these trials. Do two trials. Do a third trial if
the rates vary by more than 10%.
2. Take the reaction mixture that took the shortest time and measure the rate of this mixture in a cold
water bath, at about 16 °C. Do two trials using the solutions suspended in this water bath. Do a third
trial if the rates vary by more than 10%.
Waste Pour the left over solution into the appropriately labeled bottle in the hood.
Calculations
57
A. From the data in the first part of this experiment, determine:
1. The new concentrations of all reagents after mixing.
2. The average rate for each of the mixtures.
3. The order of reaction for C6H10O, H+ and I3 −, by comparing the various trials to one another.
4. The rate constant, k, for each trial and then the average rate constant.
B. From the data in the second part of the experiment; determine
1. The average rate for the trials at the high and low temperatures.
2. The rate constant, k, for the trials at these temperatures.
1. The energy of activation, Ea, by plotting ln k vs.
C. Given the following heats of formation:
1. Calculate the heat of reaction for the iodination of cyclohexanone.
2. Construct an energy vs. reaction pathway diagram (plot) showing the relative energy of the starting
materials, the products and the activation energy.
3. In the reaction mechanism shown below, the second step is the slow step. Show that this
corresponds to the rate law determined in lab.
58
59
Unit–8
DETERMINATION OF MOLECULAR
WEIGHT OF AN ORGANIC SAMPLE
USING RAST’S MICRO-METHOD
60
EXPERIMENT 1
DETERMINATION OF MOLECULAR WEIGHT
OF AN ORGANIC SAMPLE USING RAST’S MICRO-METHOD
INTRODUCTION
1. BACKGROUND Colligative properties are those properties of solutions that depend on the number of dissolved particles in
solution, but not on the identities of the solutes. For example, the freezing point of salt water is lower than
that of pure water, due to the presence of the salt dissolved in the water. To a good approximation, it does
not matter whether the salt dissolved in water is sodium chloride or potassium nitrate; if the molar
amounts of solute are the same and the number of ions is the same, the freezing points will be the same.
For example, AlCl 3 and K3PO4 would exhibit essentially the same colligative properties, since each
compound dissolves to produce four ions per formula unit. The four commonly studied colligative
properties are:
1. Freezing point depression,
2. Boiling point elevation,
3. Vapor pressure lowering, and
4. Osmotic pressure.
Since these properties yield information on the number of solute particles in solution, one can use them to
obtain the molecular weight of solute.
OBJECTIVE When one mole of a solute is contained in 1 kg of solvent, the melting point of that mixture is lowered by
a characteristic amount. That amount is called themolal freezing point depression constant, Kf. This can
be written as a formula:
M =
Where:
Kf = Molecular depression constant for solvent
w = Mass of solute
W = mass of solvent
∆T = Depression of melting point
M = Gram molecular weight of solute
APPARATUS Test tube, Beaker, Thermometer, Glass rod, Retort stand
CHEMICALS Camphor, Naphthalene.
61
PROCEDURE 1. Put a little pure camphor onto a clean watch glass. Now take a melting point tube sealed at one end,
and tamp it into the solid – open end down. Now invert the tube and tap it lightly on the bench to
get the solid to the end of the tube. While this method works with most dry solids, camphor has a
peculiar tenacity and it may have to be poked to the bottom of the tube with a wire.
2. The tube is now attached to a thermometer with a rubber band. The assembly is put into a thiele
tube full of oil. The tube is designed so that the hot oil will circulate evenly.
3. At the melting point the solid will become clear liquid. Record the melting point of the pure solvent
(camphor). Record the temperature when the last crystal melts.
4. Measure the melting point of a mixture of camphor and your unknown sample. A homogeneous
solution of the unknown and camphor in known concentration. weigh some camphor and some
unknown, melt them together and find the melting point of the mixture.
5. Measure the mass of an empty test tube. Add about 1 gram of camphor and record its mass. Add
about 0.1 gram of the solute (naphthalene) and record its mass. The contents of the test tube are
melted by placing it in an oil bath preciously heated to 180-185 degrees. The liquid must not be
heated more than one minute or the camphor will sublime from the solution.
6. The tube is taken from the bath, propped up in another beaker and allowed to cool. Some of the
solid is scraped from the tube and powdered on a watch glass. It is convenient to use the bottom of
a test tube as a pestle. A little of the solid is put into a melting point tube as before. The tube is
attached to the thermometer and the melting point is obtained. Record the melting point of the
mixture.
CALCULATIONS Melting point of pure camphor (T1): ..............................................................................................................
Melting point of mixture (T2): ........................................................................................................................
Depression in freezing point (∆T = T2-T1): ...................................................................................................
Mass of empty test tube (t.t): ..........................................................................................................................
Mass of t.t + camphor: ....................................................................................................................................
Mass of t.t + camphor + naphthalene: .............................................................................................................
Weight of camphor (W): .................................................................................................................................
Weight of naphthalene (w): ............................................................................................................................
M =
62
63
Unit–9
STUDY OF SAPONIFICATION OF ETHYL
ACETATE WITH SODIUM HYDROXIDE
AT EQUAL CONCENTRATION OF EASTER
AND ALKALI AND DETERMINATION
OF RATE CONSTANT (K) USING
INTEGRATED RATE LAW
64
EXPERIMENT 9
STUDY OF SAPONIFICATION OF ETHYL ACETATE WITH SODIUM
HYDROXIDE AT EQUAL CONCENTRATIONS OF ESTER AND ALKALI AND
DETERMINATION OF RATE CONSTANT (K) USING INTEGRATED RATE LAW
INTRODUCTION
1. BACKGROUND
To determine the rate of a given chemical reaction, it should be determined how fast changes are
occurring in the concentration of reactants or products. In general, if a reaction occurs A → B, the initial
substance A and substance B did not exist. After some times, the concentration of B will increase while
the concentration of A will decrease. Rate law can be determined by conducting a series of systematic
experiments on the reaction A + B → C, to determine the reaction order with respect to A. the
concentration of A is fixed while B concentration varied then the rate of the reaction is determined on the
concentration variation. For determining the order of the reaction B, the concentration of B is fixed while
the concentration of A is varied then measured the rate of the reaction on the concentration variation
Rate of reaction: The rate of reaction is defined as the change in the number of molecules of reacting
species per unit volume per unit time. It is also defined to be proportional to the concentration of reacting
species raised to a certain power called the order of reaction. It is usually taken as the rate at which the
reactant disappear or the rate at which the product is formed. The rate at which the reactant ‘a’ is
disappearing is proportional to its concentration at any instance,
i.e. rate (r)α (a - x)
Rate (r) = k (a - x)
Where k = rate constant
The concept of rate of reaction is very important in evaluating chemical reacting systems. It is the core
factor in the development of performance models to stimulate reactor functional parameters.
Factors which determine rate of reaction 1. Availability of reactants and its surface area. The greater the surface area of a solid, the greater the
rate of reaction.
2. Concentration: increase in concentration increase the rate of reaction
3. Pressure: increase in pressure results in an increase in the rate of reaction, if the reactants and
products are gaseous
4. Temperature: increase in temperature increase rate of reaction
5. Catalyst: The presences of a catalyst generally increase the rate of reaction. There are however,
negative catalysts that lower the rate of
Rate constant: Rate constant, k quantifies the speed of a chemical reaction. For a chemical reaction
where substance A and B are reacting to produce C, the reaction rate has the form:
Reaction: A + BC
r = k[A]m[B]
n
Where K is the rate constant that depends on temperature
A is the concentration of substance A in moles per volume of solution assuming the reaction is taking
place throughout the volume of the solution.
65
Rate constant is the rate of reaction when the concentration of each reaction is taken as unity. That is why
it is also known as specific reaction rate. It is a measure of the rate of reaction, so the greater the value of
the rate constant, the faster the reaction. Each reaction has a definite value of the rate constant at a
particular temperature and the value of the rate constant for the same reaction changes with temperature
and the values do not depend upon the concentration of reaction but depend upon order of reaction.
OBJECTIVE The purpose of this experiment is to show the reaction of ethyl acetate saponification by hydroxide ions :
CH3COOC2H5 + OH- CH3COO
- + C2H5OH
There are two convenient methods for measuring the progress of the reaction. The consumption of the
reactant OH- may be measured by taking aliquot portions out of the reaction mixture and measuring its
concentration by titrating with standard acid solution. The second and most convenient method of
following progress of the above reaction is through the measurement of the reaction conductance of the
reaction mixture. As the reaction shows highly conducting OH- ions are replaced by poorly conducting
CH3COO- ions. This decreases the overall conductance of the reaction mixture. In order to test whether
the reaction is second order, it is convenient to work with equal initial concentrations of both the
reactants. Under this condition, the rate equation is:
=k[A]2 (1)
And the integrated form:
=kt (2)
Where [A]° is the initial concentration and [A] is the concentration at t, and k is the second order rate
constant. The equation (2) can be rearranged into the form:
k = (3)
If co, ct and c∞ are are the conductance of the reaction mixture at time 0, t, and infinite respectively, it can
be shown that
(Co-Ct) ∞ [A]°
(Ct- C∞) ∞ [A]
(C0- Ct) ∞ [A]°- [A]
Putting these proportionality relations in equation (3) and rearranging, we get
Ct = (4)
Therefore a plot of Ct vs. Co - Ct/t should give a straight line with a slope equal to 1/[A]°.k.
APPARATUS Boiling tubes, thermostat, stopwatch, conductivity meter.
CHEMICALS 1. 0.02M Ethyl acetate.
2. 0.02M NaOH solution.
66
PROCEDURE
1. Prepare 0.02 M NaOH solution.
2. Prepare 0.02 M Ethyl acetate solution.
3. Take 50 ml of NaOH solution and CH3COOC2H5 solution in two different boiling tubes.
4. Suspend both of them in the thermostat maintained at room temperature.
5. Thoroughly rinse the conductance cell with distilled water. Measure the conductance of NaOH
solution.
6. When the two solutions have attained the temperature of the thermostat, mix them together and
measure the conductance.
7. Record the conductance of the solution at regular time intervals (1 minute- 3 minute), till there are
no measurable changes.
8. Repeat steps (6) to (9) for four different temperatures, e.g. 30°C, 35°C, 40°C, 45°C etc.
Treatment of data:
Prepare conductance – time data along with (Co – Ct)/t in tabular form.
Time, t (min.) Initial conductance, Co (μS)
Conductance at time t, Ct
(μS)
Co - Ct (Co –Ct)/t
1. Draw conductance vs. time graph and calculate Co.
2. Draw Ct vs. Co- Ct/t, and evaluate the rate constant, k from the slope (1/[A]°k) of the line.